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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