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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: I
think I might actually use all 16 colors today. Oh no, this is the
most satisfying day. Whereas Tuesday was probably
the most mathematically intense, because we developed
this equation right here, today is going to be
the most satisfying, because we are going to cancel
out just about every term, leaving a homogeneous, infinite
reactor criticality condition. So we will go over
today, how do you go from this, to what is
criticality in a reactor? So I want to get a couple
of variables up over here to remind you guys. We had this variable
flux of r, e, omega, t in the
number of neutrons per centimeter squared
per second traveling through something. And we also had its
corresponding non-angular dependent term, on just
r, e, t, if we don't care what angle things go through. We've got a corresponding
variable called current. So I'll put this as flux. We have current j, r, e, omega,
t, and its corresponding, we don't care about angle form. And today, what
we're going to do is first go over this
equation again so that we understand all of its parts. And there are more parts
here than are in the reading. If you remember,
that's because I wanted to show you how all
of these terms are created. Just about every
one of these terms, except the external source and
the flow through some surface, has the form of some
multiplier, times the integral over all possible variables
that we care about, times a reaction rate d
stuff, where this reaction rate is always going to be some
cross-section, times some flux. So when you look at this
equation using that template, it's actually not so bad. So let's go through each
of these pieces right here. And then we're going to
start simplifying things. And this board's going to
look like some sort of rainbow explosion. But all that's going to be
left is a much simpler form of the neutron
diffusion equation. So we've got our
time-dependent term right here, where I've stuck in
this variable flux, instead of the
number of neutrons n, because we know that flux
is the number of neutrons, times the speed at
which they're moving. And just to check
our units, flux should be in neutrons per
centimeters squared per second. And n is in neutrons
per cubic centimeter. And velocity is in
centimeters per second. So the units check out. That's why I made that
substitution right there. And this way, everything is in
terms of little fi, the flux. We have our first term here. I think I'll have
a labeling color. That'll make things a
little easier to understand. Which is due to
regular old fission. In this case, we have new,
the number of neutrons created per fission, times chi, the
sort of fission birth spectrum, or at what energy the
neutrons are born. Over 4pi would account for all
different angles in which they could go out, times the integral
over our whole control volume, and all other
energies in angles. If you remember
now, we're trying to track the number of neutrons
in some small energy group, e, traveling in some
small direction, omega. And those have little
vector things on it at some specific position
as a function of time. So in order to figure out how
many neutrons are entering our group from fission,
we need to know, what are all the
fissions happening in all the other groups? I've also escalated this
problem a little bit to not assume that the
reactor's homogeneous. So I've added an r, or
a spatial dependence for every cross-section
here, which means that as you move
through the reactor, you might encounter
different materials. You almost certainly
will, unless your reactor has been in a blender. So except for that
case, you would actually have different cross sections in
different parts of the reactor. So all of a sudden,
this is starting to get awfully interesting,
or messy depending on what you want to think about it. There is the external
source, which is actually a real phenomenon,
because reactors do stick in those californium
kickstarter sources. So for some amount
of time, there is an external
source of neutrons, giving them out with some
positional energy angle and time dependence. So let's call this the
kickstarter source. There's this term right
here, the nin reactions. So these are other reactions
where it's absorb a neutron, and give off anywhere
between 2 and 4 neutrons. Beyond that, it's
just not energetically possible in a fission reactor. But don't undergo fission. They have their own cross
sections, their own birth spectrum. And I've stuck in
something right here, if we have summing over all
possible i, where you have this reaction be n in reaction,
where 1 neutron goes in, and i neutrons come out. You've got to multiply
by the number of neutrons per reaction. For fission, that was new. For an nin reaction,
that's just i. But otherwise, the
term looks the same. You have your multiplier,
your birth spectrum, your 4pi, your integral over stuff,
your unique cross-section, and the flux. And these two together
give you a reaction rate. I've just written all of the
differentials as d stuff, because it takes a lot
of time to write those over and over again. And then we have our
photo fission term, where gamma rays of
sufficiently high energy can also induce fission
external to the neutrons. The term looks exactly the same. There's going to be some
new for photofission, some birth spectrum
for photofission, some cross section for
photofission, and the same flux that we're using
everywhere else. Then we had what's called the in
scattering term, where neutrons can undergo scattering,
lose some energy, and enter our group
from somewhere else. That's why we have
those en omega primes, because it's some
other energy group. And we have to account for
all of those energy groups. That's why we have
this integral there. And it looks very much the same. There's a scattering
cross-section. And that should actually
be an e prime right there. Make sure I'm not
missing any more of those inside the integral. That's all good. That's a prime, good. There's also a flux. And then there was this
probability function that a given neutron
starting off an energy e prime omega prime,
ends up scattering into r energy, e and omega. So this would be the other one. And this would be r group. But otherwise, the term
looks very much the same. And that takes care of all
the possible gains of neutrons into r group. The losses are a
fair bit simpler. There is reaction of
absolutely any kind. Let's say this would be the
total cross-section, which says that if a neutron
undergoes any reaction at all, it's going to lose energy and
go out of our energy group d, e. Notice here that these are all-- these energies and omega
is our all r group, because we only care about
how many neutrons in r group undergo a reaction and leave. And the form is very simple-- integrate over volume,
energy, and direction, times a cross-section,
times a flux, just like all the other ones. Then the only difference
in one right here is what we'll call leakage. These are neutrons moving
out of whatever control surface that we're looking at. And this can be some arbitrarily
complex control surface in 3D. I don't really know how
to draw a blob in 3D. But at every point
on that blob, there's going to be a normal vector. And you can then take the
current of neutrons traveling out that normal vector, and
figure out how much of that is actually leaving
our surface ds. The one problem we had
is that everything here is in terms of volume,
volume, volume, surface. So we don't have
all the same terms, because once we have everything
in the same variables we can start to make some
pretty crazy simplifications. The last thing we did is we
invoked the divergence theorem, that says that the surface
integral of some variable FdS is the same as the volume
integral of the divergence of that variable dV. So I remember there was
some snickering last time, because you probably
haven't seen this since, was it 1801 or 1802? 1802, OK, that makes sense,
because divergence usually has more than one variable
associated with it. I'll include the
dot, because that's what makes that divergence. So we can then
rewrite this term. Let's start our
simplification colors. That's our divergence theorem. So let's get rid
of it in this form, and call it minus integral
over all that stuff. Then we'll have
dot, little j, to be careful, r, e, omega, t, d soft. So for every step, I"m going
to use a different color so you can see which simplification led
to how much crossing stuff out. And so like I said,
this board's going to look like a
rainbow explosion. But then we'll
rewrite it at the end. And it's going to look
a whole lot simpler. So now, let's start making
some simplifications. Let's say you're an
actual reactor designer, and all you care about is
how many neutrons are here. Of the variables
here, which one do you think we care the least about? Angle, I mean do we really
care which direction the neutrons are going? No, we pretty much
care, where are they? And are they causing
fission or getting absorbed? So let's start our
simplification board. And in blue, we'll
neglect angle. This is where it
starts to get fun. So in this case, we'll just
perform the omega integral over all angles. We just neglect angle here. We forget the omega
integral, forget omega there. Away goes the 4pi, because we've
integrated overall 4pi star radians, or all solid angle. Let's just keep going. Forget the 4pi,
forget the omega, forget the omega, forget the 4pi
and the omega, and the omega. Same thing here-- forget the
omega in the scattering kernel, forget it in the
flux, forget it there, forget it there,
and there as well. OK, we've now completely
eliminated one variable. And all we had to do is
ditch the 4pi and one of the integrals. What next? We're tracking right now
every possible position, every possible energy,
at every possible time. If you want to know,
what is your flux going to be in the reactor
at steady state, what variable do you attack next? Time. So let's just say this
reactor is at steady state. That's going to
invoke a few things. For one, it's going to ditch
the entire steady state term. We're going to get rid of
all the ts in all the fluxes. This shouldn't take
too long to do. I think that's all of them. And the third thing is if this
reactor is at steady state, chances are we've taken
our kickstarter source out, because we just needed
it to get it going. But the reactor should
be self-sustaining once it's at steady state. So let's just get rid
of our source term. I just want to make sure
I didn't miss any here. OK, next up, let's
go with green. What else do you think we can
simplify about this problem? Well, if you look far enough
away from the reactor, we can make an assumption
that the reactor is roughly homogeneous. In some cases, it's not
so good of an assumption, like very close to anything
that has a huge absorption cross-section. Now, I want to explain
the physics behind this. If the neutrons travel
a very long distance through any group of
materials, then those materials will appear to be roughly
homogeneous to the neutrons. If, however, the neutrons
travel through something that's very different from the
materials around it, then that homogeneous
assumption breaks down. So in what locations in a
nuclear reactor do you think you cannot treat the
system as homogeneous? Where do the
properties of materials suddenly change
by a huge amount? Yeah, Luke? AUDIENCE: Control rods. MICHAEL SHORT:
Control rods, right, so let's say it's
bad for control rods. Where else? How about the fuel? All of a sudden, you're
moving from a bunch of structural materials
where sigma fission equals 0, to the fuel where sigma fission,
like you saw on the test, can be like 500 barns,
which even though it's got a very small
exponent in front of it, 10 to the minus 22
centimeters squared, it's still pretty significant. So this assumption breaks
down around the control rods and around the fuel. But we can get around this. Let's analyze the simplest,
craziest possible reactor, which would be a molten
salt fueled reactor. It's just a blob
of 700 Celsius goo that's got its fuel, coolant,
and control rods all built in. So if we assume that the
reactor is homogeneous, which is a pretty good
assumption for molten salt fueled reactors,
because the fuel's dissolved in the coolant. And it builds up its own
fission product poison. So it's got some of its own
control rods kind of built in. Usually, we'll have other
extra ones too, but whatever. Then we can start to
really simplify things. If we get rid of any
homogeneity assumptions, we cannot necessarily get
rid of the r in the flux, because even if the reactor's
homogeneous it still might have boundaries. So you might be able
to approximate it as just a cylinder or a
slab of uniform materials. But if we were to get rid of
the r's in the flux term, that would mean that as we graph
flux as a function of distance, it would look like that,
including infinitely far away from the reactor. Now, is was that true? Absolutely not, so I don't want
to leave that up for anyone. We'll fill in what these graphs
look like a little later, just leave them there for now. We can get rid of some
of the other r's though, like these cross sections. If the reactor is
actually homogeneous, then the cross section
is the same everywhere because the materials
are the same everywhere. So we can get rid of the
r's here, the r's here, and there, and there. And that's it, I think. I don't think I
missed any, good. Next up-- if this
reactor is homogeneous, then does it really
matter at which location we're taking this balance? Does it really matter
which little volume element we're looking at? We say these equations are-- we'll call them volume
identical, which means if this same
equation is satisfied at any point in the
reactor, we don't need to do the volume integral
over the whole reactor. It's not like it's going
to change anywhere we go. So forget the volume integrals. Hopefully, you guys see
where I'm going with this. And I've never tried teaching
it like this rainbow explosion before. But I'm kind of excited
to see how it turns out. So already like 2/3 of the stuff
that we had written are gone. What's the only variable
left that we can go after? What's the only color left
that I haven't really used? Energy, so we can make a couple
of different assumptions. This equation as it is not yet
really analytically solvable, because a lot of these
energy dependent terms don't have analytical
solutions, or even forms like the cross sections. But we can start
attacking energy. Hopefully, this is
different enough from white. Yeah, is that big enough
difference for you guys to see? Good, OK, we can start doing
this in a few different ways. I want to mention what they are. And then we're going
to do the easiest one. So the way it's done for real,
like in the computational reactor physics
group, is you can discretize the energy into a
bunch of little energy groups. So you can write this equation
for every little energy group, and assume that along
this energy scale, ranging from your maximum energy
to probably thermal energy-- 025, let's do this
clearly with thick chalk. There we go. You can then discretize into
some little energy group. Let's say that's egi, that's egi
plus 1, and so on, and so on. And depending on the type of
reactor that you're looking at, and the energy
resolution that you need, you choose the number of
energy groups accordingly. Does anyone happen to know
for a light water reactor, how many energy groups
do you think we need to model a light water reactor? The answer might surprise you. It's just two, actually. All we care about-- so let's say this would
be for the general case. All we care about for
a light water reactor is, are your neutrons thermal? Or are they not? Because the neutrons
that are not thermal are not contributing
to fission that much. They are just a little bit. And you can account for those. But pretty much, they're not. Once the neutron slowdown down
to get thermal, in the range from, let's say, about an
EV to that temperature-- took a surprising amount of time
to write with sidewalk chalk-- then you've got things that
are about 500 or 1,000 times more likely to undergo fission. And so all you care about is
the neutrons are all born. They're all born
right about here. And they scatter,
and bounce around. And you don't care,
because they're just in this not thermal region. And when they enter
the thermal region, you start tracking them,
because those are the ones that really count for fission. And if you actually look up the
specifications for the AP 1000, this is a modern reactor
under construction in many different
places in the world. When you see, how do they
do the neutron analysis? Two group approximation. So this isn't just
an academic exercise to make it easier for
sophomores to understand. This is actually something
that's done for real reactors. So if you ever felt like
I'm making it too simple, no, no, no, I'm simplifying
it down to what's really done. And I will get you
that specification so you can see what
Westinghouse says, like, this is how we
design the reactor. We made a two group
simplification in many cases. So you can discretize. You can forget it, which we're
going to call the one group approximation. Or you can-- let's say two group
is the other one that we're actually going to tackle. We're going to do this
one, forget energy. But we're not really
going to forget energy, because you can't
just pick an energy, and pick a cross-section,
and say, OK, that's the cross section
we're going to use. If most cross sections
have the following form-- if this is log of energy,
and this is log of sigma, and it goes something like
that, what energy do you pick? Go ahead. Tell me. Which energy do you pick? Anyone want to wager a guess? AUDIENCE: The ones before
or after the big squigglies. MICHAEL SHORT: The ones before
or after the big squiggles. I don't think that's correct,
because if you do it this, then you're going away
underestimate fission. If you do it here, you're going
to way overestimate fission, or whatever other
reaction you have. We didn't say which
reaction this is. The rest of you who are
silent and afraid to speak up, you're actually correct. I wouldn't actually pick
any single value here. What you need to do is find some
sort of average cross-section for whatever reaction
that accurately represents the
number of reactions happening in the system. And in order to do that, you
have to come up with some average cross-section for
whatever reaction you have by integrating over your whole
energy range of the energy dependent cross-section
as a function of energy, times your flux de over-- does this look
familiar from 1801 or 2 as well, what's the average
value of some function? Little bit? Well, we'll bring
it back here now. So retrieve it from cold
storage in your memories, because this is how actual
cross sections are averaged. For whatever energy
range you're picking-- I'm going to make this
a little more general. I won't say zero. I'll just say your minimum
energy for your group. So now, this equation is general
for the multi group and the one group and two group method. For whatever cross-section
you want to pick and whatever energy
range you're looking at, you take the actual data
and perform an average for the fast and
thermal delineation, where, let's say this is
fast and this is thermal, you would have two
different averages. Maybe this average
would be right there. You know what? Let's use white so it
actually has some contrast. So this would be one value
of the cross-section. And maybe the next average
would be right there. So you simplify this
absolutely non analytical form of your complicated
cross-section to just a couple of values. Maybe we'll call that
average sigma fast. And we'll call that
average sigma thermal. So using this analogy
and this color, we can then say,
we're going to take an average new, an average
chi, get rid of the energies, because we can perform
the same energy average integration on every quantity
with energy dependence. So all we do is we
put a bar there, ditch the energies,
ditch the energies. And let's just say that flux
is going to be what it is. Same thing here--
yeah, same thing there, and there, and there, there,
there, there, here, and here, and here. And there is a cross-section. There is an energy. There is an energy. There is a cross-section. We don't care about
those anymore. And there's a couple of other
implications of this energy simplification. What is the birth spectrum now? What's the probability
that a neutron is born in our energy group
which contains all energies? 1, OK, so forget the chi,
and that one, and that one. And what about this
scattering kernel? What's the probability
that a neutron scatters from any other energy which
is already in our group into our group, which
contains all energies? AUDIENCE: 1. MICHAEL SHORT: Yeah,
scattering no longer matters when you do the
one group approximation, because if the neutron
loses some of its energy, it's still in our energy
group, because our energy group contains all energies. So forget the scattering kernel. And forget the energy integrals. What are we actually left with? Not much. There's no green in here yet. Good, because I need
to do one more thing. There is no more green. Oh, we did green. We did time. OK, green, red, orange--
is this the orange I used? Dammit. OK, we use those. Purple, no, we've used it. Oh my God. We've used both blues. Bright yellow. Yeah? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: Yes. Chi is the fission
birth spectrum, the probability that a neutron
is born at any given energy. But because all neutrons are
born in our energy group, which contains all energies, then that
just becomes one and goes away. There's no birth spectrum,
because they're just born in our group. Does that makes sense? OK, I think I found the actual
only color, besides black on a black chalkboard, and
white which we already have, that I have left. We also have a slightly
darker shade of gray. But I'm literally out. This worked out awesome,
because there's one more thing that we want to deal with. What do we even have left? All right, what is
the one term that is not in all of the same
variables as the others? That current, that j. What do we do about that? So we're going-- sorry? AUDIENCE: The F e to e. MICHAEL SHORT: The F e to e-- so, actually, I'll recreate
some of our variables here, because there's
a lot of them. So our F of e prime
to e is what's called the scattering kernel. And that's the
probability that a neutron scatters from some other
energy group, e prime, into ours in e about de. And chi of e is the
fission birth spectrum. And just for
completeness, knew of e is our neutron multiplier,
or neutrons per fission. And I think that gives a
pretty complete explanation of what's up here. So now, let's figure out how
to deal with the current term. This is when we make one of
the biggest approximations here, and go from what's
called the neutron transport equation, which is a fully
accurate physical model of what's really going on, to
the neutron diffusion equation. And this is where
it gets really fun. You don't assume that neutrons
are subatomic particles that are whizzing about and knocking
off of everything else. You then treat the neutrons
kind of like a gas, or like a chemical. And you just say that it
follows the laws of diffusion. Again, this works out very
well, except for places where cross sections suddenly
change, like near control rods or near fuel. But for most of the
reactor, especially if we have a molten
salt fuel reactor, we can invoke what's
called Fick's law. Does this sound
familiar to anyone? Fick's law diffusion,
3091 or 5111. It's the change of a
chemical down a density or a concentration gradient. So, yeah, you've got the idea. What Fick's law says
is that the current-- or let's say the diffusion
current or the neutron current-- is going to be equal
to some diffusion coefficient, times the
gradient of whatever chemical concentration
you've got. Let me put the c in there. So right here, this
would be the current. I'll label it in
a different color. This would be your
variable of interest. Maybe c is for concentration,
or phi is for flux. Oh, that reminds me, where
are those bars on our flux? Which term did we do? Energy-- where's my slightly
lighter blue over here? All of these phi's
become capital, because we've gotten rid of
all the angular, and energy, and everything dependence. Oh, angular dependence--
neglect omega, that should be dark blue. Omega goes away. And the fluxes become capital. So many terms to keep track of. Luckily, you will never have to. And then the j
becomes a capital J. Did I miss any phi's here? No, because that one
was already gone, cool. All right, so we
can use Fick's law, and transform the current into
something related to flux. And what we're saying here
is that we're getting rid of the true physics, which is
that there's some fixed neutron current. And we're saying that neutrons
behave kind of like a gas, or a chemical in solution. And so in yellow, we can ditch
our current related term, and rewrite it. We don't have any integrals
left, as negative del squared phi. I think the only variable
left is r, not too bad. Now, we have a second
order linear differential equation describing the flow
of neutrons in the system. And so we actually
have something that we can solve for flux. I think it's time to rewrite it. Wouldn't you say? This has been fun. So let's rewrite what's left. Make sure you guys can
actually see everything there. We'll write it in
boring old white. So we have no
transient dependence. We have left sigma fission,
times flux, as a function of r. No source, and we have our
neutron nin reactions of-- oh, we forgot our
new sigma fission. Then we have our i sigma
fission from nin, times flux. Next term, we have photofission. So we have a new,
from gamma rays, times sigma fission from
gamma rays, times flux. Next up, we have-- well, last
simplification to make. We have scattering. And we have total cross-section. When we said,
forget about energy, and our scattering
kernel becomes one-- and that's light blue-- got to make one more
modification to this board. Do we care about scattering
at all anymore whatsoever? Because scattering doesn't
change the number of neutrons left. So we can then take
these two terms and just call it sigma
absorption, times flux, because if we take scattering,
minus the total cross-section, it's like saying, all that's
left if you don't scatter is you absorb. And if you remember, I'll
add to the energy pile, we said that our
total cross section is scattering, plus absorption. And absorption could
be fission and capture. And capture could be-- let's say, capture
with nothing happens, plus these nin reactions,
plus any other capture reaction that does something. So we're going to use this
cross-section identity right here with a couple
of minus signs on it. And say, well,
scattering minus total, leaves you with negative
absorption, to simplify terms. I'll leave that up there
for everyone to see. So then we have scattering
and total just becomes minus sigma absorption,
times phi of r. And we're left with-- what was that? Current, that becomes plus. There is a d missing
in there, isn't there? A yellow d, minus d. OK, and that's it. Yes? AUDIENCE: What is d? MICHAEL SHORT: d is the
diffusion coefficient right here. So we're assuming
that neutrons diffuse like a gas or a chemical with
some diffusion coefficient. And so we'll define
what that is, oh, probably next class, because
we have seven minutes. Yeah, Luke? AUDIENCE: [INAUDIBLE] MICHAEL SHORT: Uh-huh. AUDIENCE: [INAUDIBLE] MICHAEL SHORT: The
c right here, that's whatever variable
we're tracking. So let's call that flux. Or let's call it n,
the number of neutrons, because flux is just number
of neutrons times velocity. So let's say that
the concentration was the concentration of neutrons. And we just multiply by
their velocity to get flux. So it's almost like we can
say that the concentration of neutrons is directly
related to the flux. And that way, we have
everything in flux. And that's the entire
neutron diffusion equation. Yeah, this is for one group
with all the assumptions we made right here, homogeneous. What other assumptions
did we make? Steady state, and we
already neglected that. And I think that's enough
qualifiers for this. But it's directly
from this equation right here that we can
develop what's called our criticality condition. Under what conditions
is the reactor critical? So in this case,
by critical, we're going to have some variable
called k effective, which defines the
number of neutrons produced over the number
of neutrons consumed. And if k effective
equals 1, then we say that the
reactor is critical. That means that exactly the
number of neutrons produced by regular fission, nin
reactions, and photofission equals exactly the
number of neutrons absorbed in the anything,
and that leak out. So let's relabel our terms in
the same font that we did here. So this would be
the fission term. This would be nin reactions. This would be photofission. This would be absorption. This would be leakage. How many neutrons get out
of our finite boundary? And if you remember
when we started out, we said we were going to make
the neutron balance equation equal to gains minus losses. And through our rainbow
explosion simplification, we've done exactly that. These are your gains. These are your losses. When gains minus
losses equals zero, the reactor's in
perfect balance. Yep? AUDIENCE: How does leakage
come out to be negative? MICHAEL SHORT: Leakage
comes out to be negative, despite the plus sign here. And that's actually intentional. That's because
neutrons traveled down the concentration gradient. So let's say we're going
to draw an imaginary flux spectrum that's going
to be quite correct. And I'm doing all of those
features for a reason. But let's look at the
concentration gradient right here. Leakage is positive when your
flux gradient is negative. That's why the sign is
flipped right there. So a positive
diffusion term means you have neutrons leaking out
down a negative concentration gradient, because if you look at
the slope here, the change in x is positive. And the change in
flux is negative. So the slope is negative. Concentration
gradient is negative. That's why the sign
is the opposite of what you may expect. And the same thing goes
for chemical, or gaseous, or any other kind of diffusion. I'm glad you asked,
because that's always a point of confusion, is,
why is there that plus sign? That's intentional
And that's correct. Cool. Yeah, Shawn? AUDIENCE: So in
that case, if you were to explicitly
right out losses, would it be minus
absorption, plus leakage? MICHAEL SHORT: Let's
put some parentheses on here, equals
zero, and a minus. And when we say plus leakage,
we have that plus sign in there. So I'm not going to put
any parentheses up here, because that
wouldn't be correct. But what I can say is
that gains minus losses have to be in perfect balance to
have a k effective equal to 1. Does anyone else
have any questions, before I continue
the explanation? Cool. Let's say you're
producing more neutrons than you're destroying. That's what we
call supercritical. So I just did an
interview for this K through 12 outreach program. And they said, should
people be afraid when something, quote
unquote, goes critical? Sounds scary emotionally, right? And the answer is
absolutely not. If you're reactor goes
critical, it's turned on. And it's in perfect balance. That's exactly what you want. So going critical is
not a scary thing. It means we have control. If something goes supercritical,
it doesn't necessarily mean it's out of control. Reactors can be very
slightly supercritical and still in control,
because of what's called delayed neutrons, which
I will not introduce today, because we have two minutes. If a reactor has a k
effective of less than one, we call that subcritical. So it's important to note that
the nuclear terminology that's kind of leaked out
into our vernacular is not physically correct,
in the way that it's used. Words like critical are used
to incite emotions, and bring about fear. When to a nuclear
engineer critical means, in perfect control,
in balance, like you would expect,
or in equilibrium. That all sounds
kind of nice, makes you calm down a little bit. Yeah, so we can
put one last term in front of our
criticality condition. We can take either the gains or
the losses, move the equal sign and zero over a little bit, and
put a 1 over k effective here. This, then, perfectly
describes the difference between the gains and
the losses in a reactor. So if the gains equal the
losses, then k effective must equal 1. And the reactor has
got to be in balance. If there are more
gains than losses, which means if you are
producing more neutrons then you're consuming,
than k effective must be greater than 1 for this
equation to still equal zero, because this equation
must be satisfied. So if you're making
more neutrons, your k effective has got
to be greater than 1. So you have a less than
1 multiplier in front. And on the opposite side, if
you're losing more neutrons then you're gaining, your k
effective has to be less than 1 to make this equation balanced. Going along with all these
definitions right here. So it's exactly 5 of 5 of. I've given you delivered
promised blackboard of Lucky Charms. And we've hit a
perfect spot, which is the one group homogeneous
steady state neutron diffusion equation, from
which we can develop our criticality conditions
and solve this much simpler equation to get
the flux profiles that I've started to draw here. So I want to stop here, and
take any questions on any of the terms you see here. Yeah? AUDIENCE: Didn't we talk a lot
about the different energies, like the one-group, two-group,
or the discrete distributed discretized energy groups? So when we're doing
the one group, you're actually just
treating them fast together? MICHAEL SHORT: We are. That's right. AUDIENCE: To know
that, like you said, the reactors do two group
in the actual analysis. MICHAEL SHORT: So a lot a
lot of reactors, at least thermal reactors where you only
care if neutrons are thermal or not, two group is enough. When you have a one group
or a two group equation, these are fairly
analytically solvable things. You get to any more
groups than that, and, yes, they're
analytically solvable. But it gets horrible. And that's why we
have computers to do the sorts of repetitive
calculations over and over again. Once we've solved the
one group equation, I'll then show you intuitive
ways to write, but not solve, the equations for
multi group equations. Sure. Any other questions? So like I promised, we
didn't stay complex for, long because there's
basically nothing left. Yeah? AUDIENCE: What is
the 2n over 2t? Are we saying that? MICHAEL SHORT: Oh, that's
a partial derivative. Yeah, there we go. So this is saying a change
in neutron population, or the partial derivative
of n with respect to t, because n varies with
space, energy, angle, time, and anything else you
could possibly think about, equals the gains
minus the losses. I think this is
worthy of a t-shirt. If any of you guys would
like to update the department shirts to properly take
into account photofission external sources
and nin reactions, I think it would make for a
much more impressive thing, because we kind of printed
an oversimplification before. It's too bad. We definitely had
room on the shirt. There was room on the
sides, and on the sleeves. Yeah, keep going. It might have to be long sleeve. I think that would
be pretty sweet. Yeah, OK, if no one else
has any immediate questions, you'll have plenty
of time tomorrow, because the whole
goal tomorrow is going to be to solve this equation. That's only going to
take like 20 minutes. So we can do a quick review
of the simplification of the neutron
transport equation, solve the neutron
diffusion equation. If you have questions,
we'll spend time to answer them there. And if you don't, we'll move
on to writing multi group equations. And also Friday for recitation,
it's electron microscope time. So now that you guys have
learned different electron interactions with matter,
you're going to see them. So we're going to be analyzing
a couple of different pieces of materials that a couple of
you are going to get to select. And we're going to image
them with electrons to show how you can beat the
wavelength of light imaging limit, like I told you before. We're going to produce
our own X-ray spectra to analyze them
elementally, where you'll see the bremsstrahlung. You'll see the
characteristic peaks. And you'll see a couple of other
features that I'll explain too. So get ready for
some SEM tomorrow.
