## 23. Solving the Neutron Diffusion Equation, and Criticality Relations

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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: So Tuesday, we

developed the largest equation that you’ll probably

ever use at MIT. Thursday, we destroyed it down

to a rather manageable size. And today, we are going

to solve it and actually show you guys how to work out

different reactor problems all the way from simple one-group

homogeneous reactors to expanding intuitively, not

directly, from the mathematics to solve, or to pose and

solve, two-group reactor problems like the one they

did on the AP 1000 reactor where they separate the

neutrons into a fast and a thermal group. And I wanted to put

up where we left off. We have some equation describing

fission, n, n reactions, photo fission, absorption,

and leakage or diffusion as the big balance equation

for how many neutrons are there in some reactor. And I think last time, I

had some example reactors up on the board ranging

from an infinite slab with a thickness a, to a

cylinder with– we’ll call it a thickness a and a height z. And the question that we

want to be able to answer is, if we draw a graph of flux

versus x through this reactor, and in this case,

it would be r and z, what do these

functions look like? What is the form of this flux? And so today, from

this equation, we’re actually

going to solve it. So bear with me. There’ll be maybe 20 minutes

of remaining derivation, and then I’m going to

teach you how to use it. So today’s class

is going to be more like a recitation of

how do you actually use this equation

instead of getting to it. First of all, I want

to simplify things, which is going to make it– de-escalate. If we simplify things and don’t

worry about these NIN reactions and photo fission, we have the

equation that you’ll actually see in your reading. I’m going to drop them for now– and I realize I’m missing

one little gamma– because they’re

just extra terms. They were instructive in writing

the neutron transport equation because all the

terms looked similar, but now they’re just kind of

extra things for us to write. And the last thing that

isn’t in units of flux is this Laplacian operator,

for those who don’t remember. And this Laplacian operator

takes different forms in different dimensions and

different coordinate systems. For 1d Cartesian,

it’s pretty easy. It’s just double

derivative, let’s say in x. For cylindrical, it’s

significantly uglier, plus d squared over d r squared. So this is what the

Laplacian operator looks like in the case of a

finite cylindrical reactor. So first we’re going to

focus on the infinite slab case right here because

it’s a lot easier to solve, analytically. And then we’ll show

you how it would go for the cylindrical

reactor which looks a lot more like the

reactors you’ll see everywhere else in the world. So let’s start doing a

little bit of rearrangement and isolate that Laplacian,

so we can then subtract d del squared phi from each side. OK. And that cancels those out. And then we’ve got something

on the left hand, is in del squared phi,

and on the right side has all in units of flux. So then we can divide

everything by the flux. And if we then cancel

out all the flux terms, all two of

them, we’re left with something awfully simple. And the last thing we can do

is divide by d on both sides. And I’m sorry. I had to cancel that

flux because that’s the way that goes. The d’s cancel, and I’ll

redraw what we have right here. So we have minus del

squared flux over flux equals constants. And what are they actually? It’d be 1 over k times

new sigma fission minus sigma absorption over d. Everyone remember why

we’re putting these bars on the cross sections

in d and everything? Can someone just tell me? AUDIENCE: Averaged? PROFESSOR: That’s right. We’ve averaged overall energy,

so some average cross-section would be the average from our

minimum to our maximum energy of that cross-section as a

function of energy times flux, over the range itself. So by this, we’re saying, we’re

averaging the cross-section somehow over the

whole energy range. Now, I ask you guys, what sort

of functions in Cartesian space happen to have its double

derivative over itself equal to constant? AUDIENCE: Exponentials. PROFESSOR: Is what? AUDIENCE: Exponentials. PROFESSOR: Exponentials,

or sine and cosine. You’ve basically said the

same thing two different ways, yeah, because the plus and

minus exponentials can be rewritten in sines and cosines. So if we now assume that

our flux solution has to take the form of, let’s call

it a cosine bx plus c cosine, what’s the next constant? Fx. I’m sorry, that’s a sine. I don’t want to use d

again for obvious reasons, and e stands for energy. So that’s the next letter

that we haven’t really used. The flux has got to take this

shape or else this condition is not satisfied. So this is the easy way of

solving differential equations, which is guess the solution

from previous knowledge or experience or something. So based on this, if we

were to draw a flux profile, let’s say, right here was x

equals 0, one of those terms has got to go away by

reasons of symmetry. What do you think it’ll be? Which of those

halves right there is symmetric about

the x-axis right here? Sine or the cosine? AUDIENCE: Cosine. PROFESSOR: Cosine. Yep. The sines, it can be inverted

around your coordinate system, but it’s not symmetric. It’s not a mirror image. So actually, the

sine term goes away and we’ve got to have

some solution which looks like a times cosine b of

x because right here our d flux dx should equal 0 at x equals 0. Now, I’ve intentionally

drawn the flux not to go to zero at the

very edge of the reactor. If the flux automatically

went to zero at the edge of the

reactor, there’d be no need to shield it, right? So there are some neutrons

streaming out of this reactor. And this distance

right here is actually equal to two times the

diffusion constant. Let me get rid of the stuff

we don’t need anymore. This is what we call the

extrapolation length. And now I should

also mention, I’m not going to derive

it because I think we’ve done enough

deriving for one week, but I’ll just give you the form. This diffusion

constant can actually be expressed in terms

of other cross sections where this mu nought is what’s

known as the average scattering angle cosine. And it approximately

equals 2/3 times the average atomic

mass of whatever it’s scattering off of. So this d that I’ve

just introduced from some physical

analogy actually has an expression from cross

sections and the material properties that you can look up. So we’ve now turned it

into some sort of condition where everything can be looked

up in the Janis library, or whatever cross

section library you have, plus whatever number

densities you actually have in your reactor. So the picture’s starting to

get very real and very physical. So now, if we assume that phi

takes the form of a cosine bx, let’s plug it in. So if we rewrite

this expression– I think I’ll need another

color for a substitute– so we’ll take a cosine

bx and stick it in there and stick it in there. And we’ll end up with,

minus del squared cosine is going to look like

a times b squared cosine bx over a cosine

bx equals those constants. It’s starting to

get very simple. Keep the bars on there because

they’re quite important. And if we cancel things out,

the cosine bx’s go away, the a’s go away, and

we’re left with whatever is inside that cosine. B squared equals a bunch

of material properties. There’s no information in

here about the geometry of our reactor. There’s only the

material properties, whereas over here,

this constant b, I’m going to add a little g to

it which stands for geometry. And we’ve now set

up a condition where if you know the geometry and

the materials in the reactor, you can solve for the final

unknown which is its k effective. How critical or not

is this reactor? So what would b have to be

to make this cosine valid? So what would bg equal if we

have the form of this flux like so? I’ll give you a hint. If the cosine goes to zero

right here at some reactor length, a over 2, plus some

extrapolation length, 2d, then how do we make the cosine

equal to zero at this point? What does bg have to equal? AUDIENCE: Pi halves over 2d? PROFESSOR: Quite close. Bg would have to equal pi over

a over 2 plus 2d, such that when you substitute– or over 2, right, yeah. Oh, yeah, OK– so that when you

substitute x equals a over 2 plus 2d into there, this

cosine evaluates out to zero. Does that makes sense? Cool. So now, we have bg. There’s nothing

here but geometry plus a little bit of

extrapolation length right here. Yeah? AUDIENCE: Shouldn’t

there be an over 2? PROFESSOR: Oh, yeah. There should be an

over 2, so that when you plug in x equals a over

2 plus 2d, you get pi over 2, and cosine of pi over 2 is 0. Yep. Cool. So now we have our bg. So we substitute

that back in there. Let’s just continue to call it

bg since we have it over there. And we have this much simpler

expression, bg squared equals 1 over k mu sigma fission plus

sigma absorption over d. Then, where’s our

rearranging color? We can multiply everything by d. And let’s see. That should have been a minus. Copied myself over wrong. The d’s go away. You can then add sigma

absorption to each side. And then those go away. And then, you can

multiply everything by k, and those go away. And finally, divide everything

by sigma a plus d bg squared. And what we’re left with is

our criticality condition. Our k, our criticality,

is pretty simple, nu sigma fission over

absorption plus leakage. So finally, after all

that derivation, we’ve arrived at some

intuitive result. Remember we said that k,

the criticality, or the k effective, is the ratio of

gains to losses of neutrons. So that’s exactly what we have. The only gain mechanism

right here is by fission, and the loss

mechanisms are either absorption by the stuff

in the reactor or leakage outside of the reactor. And so this is

actually how you tell when the reactor’s

in perfect balance, is if this condition is

satisfied, and if it equals 1, then the reactor is critical. So now, we can start

to play with this. And let’s say we started

off with a critical reactor and all of a sudden, we were

to boost the absorption. What should happen

to k effective? If you start observing

more, right, it should go sub-critical. Now, mathematically speaking,

that means this denominator gets larger. So this ratio gets smaller. And therefore, k effective

has to go down, as well. Now that’s an easy case. Let’s start exploring some

of the more interesting ones. Let’s say you raise the

temperature in the reactor. Do we necessarily know

what’s going to happen next? Let’s work it out. So most cross sections, when

you raise the temperature, will actually go down in

value due to a process called Doppler broadening that

you’ll learn about in 22.05. But suffice to say for now

that cross sections tend to go down with temperature. The most important reason why

is because a cross section is a number density times a

microscopic cross section, and if the temperature goes

up, then the density goes down, and the number

density goes down. And if the number

density goes down, the macroscopic

cross-section goes down. The atoms just spread

out from each other. So regardless of what happens at

the microscopic cross-section, which I’ll leave to Ben and

Cord to teach you next year, we know that the macroscopic

cross-section goes down because it gets less dense. So let’s try and

work out now, what would happen to k effective? So what will happen to nu

if we raise the temperature? Nothing, let’s hope. What happens to sigma fission? This goes down a bit. But what about sigma absorption? Sigma absorption is

going to go down. Does bg change? Has the geometry of

the reactor changed? Probably not. Might have thermally

expanded by a few nanometers, but let’s just say, it

doesn’t change at all. What about the

diffusion constant? Let’s work that out. Sigma total is going to go down. Sigma scattering is

going to go down. So probably what’s

going to happen is, this diffusion

constant is going to go up, which means that

if the atoms spread out more, neutrons will move

farther, on average. Hopefully, that makes

intuitive sense, because if the

cross-sections go down, then a neutron can move farther

before an average interaction. So the diffusion constant

is probably going to go up. Which way does k effective go? You’re correct. No one said anything because

you can’t really say anything. So it depends on

the relative amounts that these increase or decrease. So depending on what you

choose for your materials, you can have what’s called

positive or negative temperature feedback, which

means in some conditions or scenarios, what

you want to happen is that if the temperature goes

up, k effective should go down, but not necessarily so. Depending on what you use, you

can actually have situations where raising the temperature

raises k effective, and that is some seriously bad

news and is actually outlawed. You can’t design a reactor

with positive temperature coefficients. So this is the first little

taste of reactor feedback is, now that we’ve written

this criticality condition, we can start to explore

what happens when you start probing the reactor. So let’s say, what happens

if you just add more reactor? In this case– where’s my green? All the way over there– without changing the

materials, what happens when you make the reactor bigger? What increases, decreases

or stays the same? Let’s just work it through. Does nu change? Sigma fission? No. Sigma absorption? D? How does bg change? Bg decreases, and

as you’d expect, if you add more reactor

to your reactor, the k effective should increase. And so this,

hopefully, is starting to follow some

intuitive pattern. With a given criticality

condition, in some situations, you can work out, will the

reactor gain or lose power? Speaking of, where’s the power? Where’d it go? Yeah? AUDIENCE: The kinetic

energy of neutrons? PROFESSOR: So yes,

the power comes from the kinetic

energy of neutrons, but where did the power

go in our expression? Yeah? AUDIENCE: Power’s not

dependent on criticality. PROFESSOR: That’s right. That’s exactly right. And it follows

directly from the math. This a got canceled away. It doesn’t matter. You can actually have what’s

called a zero power reactor. So the power of the

reactor and its criticality are not necessarily linked. You can have a reactor that

is critical while producing tons of power. You can also have a reactor that

is critical while producing– I won’t say zero, but an

infinitesimally small amount of power. And they actually

have built these. They’re great test

systems for testing our knowledge of neutron

physics because you’ve got a reactor that’s producing

maybe 10 watts of power. It’s easy to cool by blowing

a fan on it, let’s say. But you can still

measure the neutron flux in different places and test

how well your codes are working with a much safer configuration

than sticking probes into a gigawatt

commercial reactor. Yep? AUDIENCE: So [INAUDIBLE]

steady state reactor, how are you [INAUDIBLE] if it’s

not really at the steady state? PROFESSOR: That would push the

reactor out of steady state. Indeed, so on

Tuesday, we’re going to start covering transience,

and if k effective become something other

than one, the reactor is no longer in steady state. It’s not in equilibrium because

the gains and the losses are not equal to each other. And at that point, the

power will start to change, what you guys all saw when you

manipulated the reactor power. So since you brought it

up, does anybody remember, if we draw as a

function of time, let’s say the reactor

power was cruising along, and right at the time is now,

you withdrew a control rod. What happened when

you guys did that? Anyone, because you all did it. It went up. OK. And then what? When you stopped withdrawing the

control rod, did it level out? So everyone, tell

me what happened. AUDIENCE: It slowed down. PROFESSOR: It slowed

down the increase, but it didn’t stop going up. Kind of freaky. So this is why I had you guys

do that power ramp because just controlling a reactor is not as

simple as, remove the control rod, you remove a certain

amount of reactivity because there are

time-dependent effects due to delayed

neutrons, neutrons that aren’t immediately

released after fission that can have a large effect on

how you control your reactor. And then if you wanted

to decrease it again, let’s say you put

the control rod back into its original

position, the power would not come back to

its original position. But then, eventually, it

would start to coast down and probably go beneath

its original position at which point you have to

constantly be controlling those control rods

to keep it in what I’ll call dynamic equilibrium. You never really hit static

equilibrium unless it’s off. As I went to a seminar a

couple weeks ago and said, I don’t study biological

organisms in static equilibrium because that’s better

known as a dead organism. They’re not very interesting. But dynamic equilibrium sure

is, for them and for us. So with this process of getting

the single-group balance equations, I’d like

to generalize this to the two-group

balance equations. And this is something

you can actually use. In every case,

we’re going to say, let’s put our gains on the

left and put our losses on the right, if we want to have

this reactor in equilibrium. And now we’ll

separate our equations into the fast and thermal

regions of neutron energy. So we’ll call those f, and

we’ll call the thermal ones th. So using this model of our

neutron diffusion equation, what are the gains of neutrons

into the fast spectrum? AUDIENCE: Straight from fission. PROFESSOR: Yeah,

straight from fission. So how do we write that? This process that we’re

going through now, this is where recitation

really begins because this is how I want to show you guys

how to approach a problem, let’s say, a one-sentence

statement like, give me the flux anywhere in

a two-group reactor. This is how we go about it. So how do you equationally put

the neutron gains from fission? What terms do we have

up there right now? AUDIENCE: Your neutron

multiplication factor and your cross section. PROFESSOR: Yep. Yep. You’ll have your nu, your

neutron multiplication factor. And now, we’re actually going

to split every cross section into its fast and

thermal energy ranges because now we’re actually

splitting that energy, like we did when I drew that

crazy cross section. Let’s see, we had log of

e versus log of sigma, and they all follow

roughly that formula. And we split it and

said, if we want to draw an average

cross section, it would look

something like this. And that would be

our sigma thermal and this would be

our sigma fast. So that’s what we’re doing here. So now it gets a

little more complicated because both fast

and thermal neutrons can contribute to fission. So how do we write this

in terms of equations? AUDIENCE: [INAUDIBLE] PROFESSOR: We only

want the neutrons that are born into the fast

region, the fast gains. That doesn’t mean you don’t

have to consider where are the thermal neutrons,

because it’s mostly those thermal neutrons that,

when they get absorbed and make fission, create fast neutrons. So what we’d really need

is sigma fission fast times our fast

flux because we’re going to split every variable

into its fast and thermal parts, plus– let’s put a parentheses there– sigma fission

thermal phi thermal. So do you guys see

what I’ve done here? We’re assuming

that every neutron is born in the fast

group, where we’re cutting this off at around 1 ev. And we are assuming

that no neutrons are born below 1 ev, which

is a very good assumption. So in this case, both the

fast and the thermal fluxes contribute to creating

fast neutrons. Is there any other

source of fast neutrons? Good, because I don’t

know of one either. OK what about losses? By what mechanisms can

neutrons leave the fast group? Yeah? AUDIENCE: Aren’t they absorbed? PROFESSOR: Yeah. They can be absorbed. So how do I write that? AUDIENCE: Sigma af– PROFESSOR: af– AUDIENCE: –times the flux fast. PROFESSOR: Times the fast flux. So only neutrons in

the fast flux group will leave the fast flux

group by absorption. And what’s the other mechanism

that we had in our neutron diffusion equation? AUDIENCE: Scattering. PROFESSOR: Yeah,

actually, so that’s not in the diffusion

equation, but you are right. That’s the missing piece that is

going to be the hard part, so. Let’s add that in now. So there’s going to

be some scattering from the fast to the thermal

group, times our fast flux. So not every

scattering event will cause the neutron to leave the

fast group, but some of them will. So we have to

figure out, what is the proportion of

those neutrons that will scatter from the fast

group to the thermal group? For the case of hydrogen,

it’s pretty easy because the probability of

a neutron landing anywhere from zero to e, starting

off at energy ei, if we had our scattering

kernel, is a constant. So that’s not too hard. And then last, what other

way can we lose neutrons from the fast group? AUDIENCE: Leakage. PROFESSOR: Yep. Leakage. They can leave the

reactor, and we can write that as a d

fast bg squared flux. Make sure everything has

bars that needs them. OK. Now, using the same

sort of logic, let’s– Yeah, Luke? AUDIENCE: What’s the bg? How is that different

from [INAUDIBLE]?? PROFESSOR: It’s not. It’s the same. It is the same bg that describes

the geometry of the reactor. AUDIENCE: I guess what’s

the subscript b of g? PROFESSOR: G means geometry. Yep. And you had a question? AUDIENCE: Yeah, just in the last

flux, [INAUDIBLE] fast flux. PROFESSOR: Yes, thank you. That is a fast flux. Yep. But it’s important to note

that this flux right here is not a fast flux. We’ll get back to that soon. Now, using the

same sort of logic, let’s write the gains and

losses in the thermal group. So what is the only

source of neutrons into the thermal energy group? I want to hear from someone

who hasn’t said anything yet. So Jared, what would you say? Either Jared because I haven’t

heard from either of you. AUDIENCE: [INAUDIBLE] PROFESSOR: You did. OK, then you. I’m sorry. All right. Yeah, you said the

no power thing. Thank you. AUDIENCE: So could

you, like if something is absorbed in

the fast spectrum, jump down to the thermal? PROFESSOR: Close. I want to replace one

word in what you said. If something is blank

in the fast spectrum, it goes down to the

thermal spectrum. AUDIENCE: Scattered. PROFESSOR: Yes, scattered. Every neutron that leaves

the fast group by scattering enters the thermal group

also by scattering. And in this case, we want to

have the fast flux appear here because the number of neutrons

entering the thermal group depends on how many scatter

out of the fast group. Yeah, Luke? AUDIENCE: Would you ever

scatter up into the fast group? PROFESSOR: You’ll see. Yes. Yeah, great. I just gave something away. Yes. You can, but no,

you usually don’t. So we would consider

that once neutrons enter the thermal group,

they’re at thermal equilibrium with the stuff around

them, and up scattering is rarely a possibility. You’ll see. Yeah, quite soon actually. Don’t worry. Not like quiz you’ll

see, but you’ll see. Yeah. I’ve already got some

stuff planned out. It’s going to be a part

of the homework question. So now what loss

mechanisms do we have? AUDIENCE: Leakage. PROFESSOR: Yeah, leakage. So we’re going to have some

separate thermal diffusion coefficient because that

diffusion coefficient depends on the cross sections

which depends on the groups you’re in, times

the same geometry, times phi thermal. And what’s the only

other mechanism of loss? AUDIENCE: Absorption. PROFESSOR: Absorption. We’ve got to clear a. Why is there no scattering

from the thermal group? AUDIENCE: Didn’t you say

it was very rare to have it scatter up to the fast group? PROFESSOR: I’d say even simpler. Once you’re at the

bottom, there’s no more lower you can go. So in neutronics, when

you hit the bottom, you don’t say,

throw me a shovel. You say, you’re at the

final energy group. So now, what we’d

like to be able to do is, last thing we want to

stick in is our k effective, our criticality,

because in reality, this is kind of what we want to

know in terms of the geometry and the materials

in the reactor. So if we know what we make it

out of and how big to make it, we should be able to get

those in balance such that k effective equals 1. So the only really unknown here

besides the flux unknowns is k. And the reason I don’t care

about the flux unknowns is, they’re going

to go away soon. Yeah? AUDIENCE: Does the thermal also

have the [INAUDIBLE] over k? PROFESSOR: Absolutely,

because the k effective is on the bottom of the total

original sources of neutrons. Just like, let’s see. That was one group. Yeah. So I’d say right

now, this accounts for the production

of all neutrons, and everything else down

the chain is losses. Yeah, Monica? AUDIENCE: Do we assume that

all neutrons [INAUDIBLE]?? PROFESSOR: We know,

experimentally, that they tend to be born

between 1 and 10 mev, but since you asked, let’s

escalate the problem. And then we will de-escalate

very quickly, just to say, let’s do a thought

experiment, right? Let’s say some of the

neutrons were born thermal. What would we have to

add to this expression? There’s one variable

missing that’s not anywhere on these boards, but was

there on Thursday and Tuesday. AUDIENCE: Spectrum? PROFESSOR: That’s right. Chi, the birth spectrum. So if some neutrons

are born thermal, then we would have to

add a Chi fast here, and we would have to add

a Chi thermal to say, this is the proportion of

neutrons born fast or thermal, times nu times

sigma fission fast, phi fast plus sigma fission

thermal, phi thermal. And I’m not writing

nice because I’m just going to erase it in a second,

but to go with your thought experiment, this

is what it would look like if some of the

neutrons were born thermal. Perfectly fine thing to model. Doesn’t happen

much in real life, but great exam

question for next year. Thank you. AUDIENCE: Next year. PROFESSOR: Next year. I’m not going to give you

your own exam question. That’s just too easy for you. So for now, let’s

forget about that stuff and stick with the most

realistic situation. Ah, running out of room already. OK. That’s for next week. So let’s forget about that. What do we do next? We have two equations

and three unknowns. Interesting. Or do we really? Well, for one thing, if we

can get that top equation all in terms of one of the

fluxes, either fast or thermal, then every term is

in terms of a flux and they can all be divided out. So let’s take one

of these equations and substitute in so

that we get everything in terms of only one flux. So let’s say, the top one,

which has got the k in it, has one instance of phi thermal. So let’s isolate phi thermal

in terms of everything else. So we have that thermal

equation right there. So we have sigma scattering

from fast to thermal times fast flux equals two things

times phi thermal, which is d thermal bg squared plus

sigma absorption thermal. We’re actually

not that far away. So all we do is, we

divide each side– where’s my simplifying color– substitute. That’s not it. Rearrange. Divide everything by this

stuff, and those cancel out. And we’re left with an

expression for phi thermal which we can now plug

into that top equation. So we’re like one step

away from the final answer. There, everything’s

still visible. And so now we end up

with 1 over k times nu sigma fission

fast, fast, plus sigma fission thermal

times this expression, sigma scattering fast

to thermal, phi flux over d thermal. I don’t usually spend this

much of the class with my back to you, but this is pretty

mathematically intense, so I apologize for that. And equals sigma absorption

fast, fast flux, plus d fast bg squared fast flux. That’s not a bar. That has one. That has one. That does. That’s good. OK. Now every single term here

is in terms of fast flux. So we can just cancel them

from every single term here. And now we’re left

with an expression for k effective that’s just in

terms of material properties and geometry for the

two-group problem. We’re only one step away. So if we multiply everything

by k and divide everything by this stuff, we’ll just

have a sigma absorption plus d fast bg squared. That would equal k. And just like that is

the criticality condition for a two-energy

group homogeneous reactor of any geometry. All that matters to

define the geometry is, what’s this bg squared? So this case works for

an infinite slab reactor. It works for an actual

right cylindrical reactor. You just have to

sell for or look up the correct buckling

term, this bg squared, which I’ll tell

you now, we refer to as buckling or

geometric buckling, and you’ve got the

solution to this. Let’s just check to see

what we actually have here. We have nu material

property, material property. All of those are material

properties except for the bg’s. So this tells you how to

design a reactor, physically, and in terms of which

materials to make sure that it’s critical. And if we look at what

this looks like here, again, it’s a ratio of gains

to losses because eventually, the losses right here, these are

the losses from the fast group. These are the losses

from the thermal group. These are the gains

in the fast group, noting that some of the

neutrons born in the fast group scatter out of

the thermal group, but don’t leave the reactor. So again, it turns out to

a gains over losses ratio. And there you have it. So I want to stop at 10 of– Yeah. AUDIENCE: Did we drop

the scattering term from the fast equation? PROFESSOR: It

should be– did we? Yeah. Let’s stick it in right here. So we’ll just also

stick it here. There. And that flux goes away because

it was in terms of everything. Yeah. There we go. Thank you. OK. But again, this

represents losses on the bottom, gains on the

top, just like any other k effective. So I wanted to stop here

at 10 of, 5 of, and answer any and all questions

you guys have about going from the

neutron transport equation all the way to something that

you could solve, and then start to play

around with to say, what happens if I

switch isotopes? What happens if I

raise the temperature? What happens if a chunk

falls off of the reactor and it gets smaller? Yeah. AUDIENCE: We got the

equation for [INAUDIBLE].. PROFESSOR: Yes. AUDIENCE: [INAUDIBLE]? PROFESSOR: Somewhat. We can assume that for

considerably long enough times, and to a neutron, a long

time could be like seconds, that the time and the

spatial form of the flux are separable which is

something that we’ll talk about on Tuesday. But, if you remember, one

of the major assumptions we made in the neutron transport

equation was steady state. We got rid of any

transient effects. We’ll bring them back, now that

we have a way simpler case, on Tuesday. Yeah, Luke? AUDIENCE: [INAUDIBLE]

step, the plus scattering– PROFESSOR: From fast to thermal. AUDIENCE: Is that also supplied

by the sigma [INAUDIBLE]?? PROFESSOR: Where is that going? AUDIENCE: It must be

in the denominator, right, because it was

over on the right side? PROFESSOR: Let’s see. Oh, yeah, we divided by all

the stuff on the right side, didn’t we? OK. So that shouldn’t be there. But it should be there because

we divided by everything on the right side. Let’s just check that

really carefully. So it should have been– no, that’s the thermal one. So we’re not

worrying about that. Yep. So it would just end up here. Yeah. Good point. Cool. Let’s talk a little

bit about what I’d want you guys to be

able to do with this. So what would I want you to

be able to do on the homework and on an exam? With the neutron

transport equation, recite it from memory. Well, not really. But if I were to give you the

neutron transport equation, I’d maybe want you to explain

what some of the terms mean, or tell me how you

would get the data, or explain one of the

simplification steps and justify why you think

it’s OK because we actually wrote out the justification

for every step on the board. Or explain, for example, what’s

the physical reason that we can solve the neutron transport

equation with this diffusion approximation? And in which regions does

that approximation break down? So can anyone tell

me, from yesterday, where is the diffusion

equation a bad approximation of the flux? Yep. AUDIENCE: Near the

control rods or the fuel. PROFESSOR: Near the

control rods or the fuel, or anywhere else where cross

sections change all of a sudden because diffusion describes

long distance steady state solutions across

places, and where things change drastically,

diffusion breaks down. Because we assumed here

that the neutrons behaved like an ideal gas or

some chemical species with no neutron to

neutron interactions, because the mean free path

length for those interactions is like, what did we say,

10 to the 8th centimeters, so a megameter? Yeah. I love using those

sorts of terminologies. 1,000 kilometers

before a neutron would hit another neutron. Or I might ask

you to, let’s say, reduce the neutron

diffusion equation and come up with a simple

criticality condition. Or let’s say, if you were

to make a physical change to the reactor,

tell me if you think it would go more

or less critical, and what would happen next? Or I could give you a

different physical situation, like the up scattering

scenario, which I will, and ask you to pose and

maybe solve these equations, or at least get forms of

the criticality condition. I’m not going to ask you to

get tons of flux equations because that’s all

22.05 is about, is doing this sort

of neutron physics. But I want to make sure that you

walk into that class prepared. Plus we’ve been kind

of heavy on the– you know, this

class, the name of it is Intro to Nuclear Engineering

and Ionizing Radiation. And so far, we’ve been

pretty heavy on the ionizing radiation and physics. So this is where the

engineering comes in. Assuming you have some

material properties, you can now pick them

to create a reactor in perfect equilibrium. Yeah, Kristin? No? So did anyone else have any

questions about the material or about what I might

ask you to do with it? Yeah. AUDIENCE: You said

this equation would hold for any geometry just

based on [INAUDIBLE] neutrons. PROFESSOR: Yep. So all that you

would do differently is, right here, when we had

that Laplacian operator, we took the one-dimensional

case of an infinite reactor in finite and one

dimension, which meant the Laplacian operator

is just double derivative of x. But you could pose the equation

in cylindrical coordinates and say, well, let’s say now

you had an infinite cylinder reactor, you

wouldn’t necessarily have sines and cosines that

would satisfy this relation. Anyone happen to

know what you’d have? The sines and cosines of

the cylindrical world, called Bessel functions. So these are the sorts of, in

cylindrical geometry equations, that behave similar

to sines and cosines with kind of regular

routes and that you can describe in a similar way. But I’m not going to

get you guys into that. I’ll just say, OK,

there exists solutions that you can look up in

the cylindrical case. And I would not make you

derive them by hand because, what’s the point? Again I’m not here

to drill your– can you do the same math

over and over again? I want to make sure that you

can intuitively understand, what’s a k effective? In a sentence, it’s

gains over losses. What happens when you push

that out of equilibrium? Or what physical situations

could push that out of equilibrium? So any other questions

for you guys? Yeah. AUDIENCE: Just curious

on the cylindrical graph we have there, what would

the graph in flux look like? PROFESSOR: It would

look pretty similar. In r, it would kind of

come down like that. It would always be symmetric

about the center for symmetry arguments, and in z, it

would look kind of like that. And so actually in the end,

the form of flux in r and z comes out as the

first Bessel function. Let’s say, that’s a times the– what would you call it — times cosine of this distance

is z, so pi is z over– I’m going to have

to add a subscript. And there’d be some

constant a in there. So what you can assume for

multi-dimensional reactors is that the dimensions

are separable. So the r part is solved

separately from the z part. And that comes right from the

Laplacian operator right here. If you assume that

some flux in r and z can be written like the

r part as a function of r times the z part

as a function of z, then the solution gets a

lot easier to deal with. But this is not

something I’d ask you to do in any coordinates

but Cartesian because those are more intuitive, and you’ll

get plenty of the other stuff later. AUDIENCE: What’s the name

of those functions again? PROFESSOR: They’re

called Bessel functions. So if you want to look up,

there’s a little bit about them in the reading, but it’s one

of the more advanced topics that I’m not going to have

you guys responsible for. Much rather it be you’d

be able to tell me what happens if you

compress the reactor or raise its temperature,

or pull out a control rod, or raise the pumping speed

and cool down the water, or something like that. So there’ll be plenty of

those kinds of questions on the homework to help

reinforce your intuition, as well as some of the noodle

scratches will be developing a criticality reaction

or equation for a more complex system than the one

I’ve just shown you here, but using the same methodology

and the same ideas.

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