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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: So today is going
to be the last day of neutron physics. As promised, we're
going to talk about what happens as a function of time
when you perturb the reactor, like you all did
about a month ago. Did any of you guys notice
the old-fashioned analog panel meter that said, reactor period,
when you were doing your power manipulations? We're going to do that today. And you're going to explore
that on the homework. So I'm arranging for all of
your actual power manipulation traces to be sent to you. So each one, you'll have
your own reactor data. You'll be able to describe
the reactor period and see how well it fits our
infinite medium single group equations, which it turns
out is not very well. But that's OK, because you'll
get to explain the differences. First, before we
get into transients I wanted to talk a bit about
criticality and perturbing it. So let's say we had our old
single group kit criticality relation. And I'd like to analyze,
just intuitively or mentally with you guys, a few
different situations. Let's say we're talking
about a light water reactor or a thermal reactor,
like the MIT reactor, or pretty much all the reactors
we have in this country. What sort of things could
you do to perturb it? And how would that
affect criticality? For example, let's say you
shoved in a control rod. Let's take the
simplest scenario. Control rods in. What would happen
to each of the terms in the criticality condition? And then, what would
happen to k effective? So let's just go one by one. Does nu ever change, ever? Actually, yeah, it does. Over time, you'll start-- that nu right there,
remember, that's a nu bar, number of neutrons
produced per fission. As you start to consume
U238 add neutrons. And as you guys saw through
a complicated chain of events on the exam, eventually
make plutonium 239, which is a fissile fuel. The nu for 238 is actually
different than the nu for 239. So I don't want to say
that nu never changes. It's just that shoving the
control rods into the reactor is not going to change nu. But it does change slowly over
time as you build up plutonium. What about sigma fission? If this were a blended
homogeneous reactor or a reactor in a blender, what
would happen to sigma fission as you then shove in
an absorbing material? Does it change? AUDIENCE: No. MICHAEL SHORT: You say, no. And I'm going to add
here homogeneous. So in this case, remember if
we define the average sigma fission as a sum-- I'll add bits to it-- of each material's volume
fraction, or let's say atomic fraction, times each
material's sigma fission, if we throw nu materials
into the reactor, then this homogeneous
sigma fission does change when
we put materials in or take materials out. So you guys want to
revise your idea? AUDIENCE: Yes. MICHAEL SHORT: Yes, thank you. There's only one other choice. Now the question
is, by how much? If you put in a control rod
where let's say the control rod's sigma fission
would be equal to zero, but volume would
be equal to small. Can't be any more
specific than that. How much of an
effect do you think you'll have on sigma fission? AUDIENCE: Small. MICHAEL SHORT: Very small. So let's say a little
down arrow like that. What about sigma absorption? The volume is still small, but a
control rod by definition sigma absorption equals huge. So what do you think? AUDIENCE: It's
going to increase. MICHAEL SHORT: It's going to
increase a little or a lot? AUDIENCE: A lot. MICHAEL SHORT: Quite a bit. Now let's look at the
diffusion constant. And remember that the diffusion
constant is 1 over 3 sigma total, minus the average
cosine scattering angle sigma scattering. What do you think is going to
happen to the neutron diffusion coefficient as you throw
in an absorbing material? Something that's got an enormous
absorption cross-section is also going to have an
enormous total cross-section, because sigma total is
sigma absorption plus sigma scattering. And sigma scattering
doesn't change that much. But if sigma absorption goes
up, sigma total goes up. If sigma total goes
up, then what happens o the diffusion coefficient? AUDIENCE: Decrease. MICHAEL SHORT: Yep,
it's got a decrease. And how does inserting a
control rod change the geometry? AUDIENCE: It doesn't. MICHAEL SHORT: Very, very close. Yeah, you're right. The control rod better
not change the geometry, but what I do want
to remind you of is that this buckling term
includes-- let's say, this was a one
dimensional infinite slab Cartesian reactor. That little hot over there
means we have some extrapolation distance. Remember, if we were to
draw our infinite reactor with the thickness
A and we wanted to draw a flux profile
on top of that, it would have to be
symmetric about the middle. And let's say we had our axis
of this is x and this is flux. Flux can't go to zero right
at the edge of the reactor, because that would mean that
no neutrons were literally leaking out. So there's going to be
some small extrapolation distance equal to
about two times the diffusion coefficient. So the geometric
buckling is actually pi over the reactor
geometry, plus 2 times the diffusion coefficient. And if the diffusion
coefficient goes down, but it's also very,
very small compared to the geometric buckling, how
much does the buckling change and by how much? And in what direction? AUDIENCE: It increases
very slightly. MICHAEL SHORT:
Increases very slightly. So the buckling might
increase very slightly. What's the overall net
effect on k effective? AUDIENCE: It goes down. MICHAEL SHORT: Should
go down, you would hope. If you put a control
rod in, it should make k effective go
down, because there's a little decrease here. Things kind of cancel out there. But the big one is putting
an absorption material, like a control rod in, should
make k effective go down. And that's the
most intuitive one, but you can work out one
term at a time what's generally going to happen. So let's now look at
some other scenarios for the same
criticality condition. I'll just rewrite it so that
we can mess it all up again. Now we want to go for the case
of boil or void your coolant. And now we're getting into the
concept of different feedback mechanisms. We've already talked
once about how raising the temperature
of something tends to increase
cross-sections in certain ways. But now let's say,
what would happen if you boil your coolant? If things got really hot and
the water started to boil. What do you want to
happen to k effective? You want it to increase? AUDIENCE: Decrease. MICHAEL SHORT:
Decrease, thank you. You want it to decrease, or
else you'd get a Chernobyl. And we'll talk about how that
happened in a week or two. So now let's reason
through each one of these. Let's assume that
nu doesn't change when you boil the coolant. What about sigma fission
of the whole reactor? You're taking a little bit of
material out of the reactor by taking liquid water,
which is fairly dense, and making it gaseous
water, which is less dense. So overall, there are more
fissile atoms in the reactor proportionately when the
coolant is boiled away than when it's not. So what happens
to sigma fission? The average sigma
fission for the reactor will go up ever so slightly. Probably not enough to matter. What about sigma absorption? If the coolant disappears. AUDIENCE: It goes down. MICHAEL SHORT: Yeah,
water is an absorber. Hydrogen and oxygen--
really just hydrogen-- have some pretty non-negligible
absorption coefficients. And if those go
away, then you're losing a bit of
absorber, aren't you? Actually it's interesting. Oxygen is the lowest thermal
cross-section of any element. So we can treat it as
pretty much transparent. Now how about the
diffusion coefficient? We've got the formula
for it up there. If all of a sudden
your neutrons don't have much to moderate from-- there's not much to
moderate your neutrons. Yeah. AUDIENCE: Your scattering
just disappears. MICHAEL SHORT: Your scattering
just disappears, right? But so does some of your
total cross-section. So chances are,
those neutrons are going to go farther before
they undergo any given collision because there's
no water in the way. So you'd expect neutron
diffusion to go up. And what about
geometric buckling? Diffusion goes up, then
the geometric buckling-- I'm just going to
make it really small. But the net effect here, once
again, k effective goes down. We didn't talk about anything to
do with the actual temperature effects on the cross-sections. This is just a density
thing on the coolant itself. So let's now look at that. What about if you
have some power spike, raised fuel temperature? I'll write it again, so
we can mess it up again. So let's say you raise
the fuel temperature. And that's going to cause
every cross-section effectively to increase if you're doing
this average scenario. Let's talk a little
bit about why. It's not as simple
as just saying, the cross-sections go up. So let's say we had two
different temperatures, cold and hot. So this would be your
sigma fission cold. And this would be your
sigma fission hot. For cold, sigma fission
looks something like that. And as the temperature
goes up, these resonances, which I'll
just label right here-- resonances being
specific energy is where the absorption suddenly
goes up, suddenly goes down will actually
decrease in height. But they'll start
to spread out more. That's about as well as I
can draw it very crudely. And same thing goes not
just for sigma fission, but for sigma anything,
including absorption, including total whatever you want. And so if your goal is
to get your neutrons from the fast
region where they're born into the thermal region
where you get fission, broadening these cross-sections
makes it more likely that if the neutron loses
any amount of energy, it's going to hit one of
these big resonance regions and get absorbed or taken
away before it gets a chance to go to the fission region. So what this is
really going to do-- it's kind of funny to say
it in terms of a one group criticality relation, but
your fission cross-section is actually going to go down. One reason is that the fuel
physically spreads out. And so just from the
density modification, you're not going to get as much. But then you've
also got that effect of increasing fission from these
resonance regions spreading out. The question is, which
one is a bigger effect? Can't answer that with
a simple statement. You'll go over a lot
more of that in 22.05 when you talk about what
actually defines a resonance region, how do you
calculate them, and how do they Doppler broaden
or broaden with temperature. How about sigma absorption? AUDIENCE: It goes down. MICHAEL SHORT: Yeah,
sigma absorption, it's going to go down because
things spread out. But it might also go up because
the cross-sections spread out, or the resonances spread out. What's really going to
happen though is the reactor atoms are effectively
spreading themselves apart. The coolant's less dense. The structural materials
in the fuel and everything are still there. They're less dense,
but there's not fewer of them in the reactor. But there is going to be
less coolant in the reactor, because it has the
ability to sparsify or get less dense, and kind of
squeeze out the inlet and outlet of the reactor. So what's really
going to happen here is, we know diffusion
is going to go up, which might cause a
corresponding change in buckling. And the net effect,
as we would hope, k effective would go down. And so what we've
talked about now here is directly
controlling reactivity with control rods, what's
called a void coefficient, where you actually want to have
a negative void coefficient. So if you boil your
coolant too much, k effective should go down. And that's one of the
mechanisms that a light water or a thermal reactor can
help stabilize itself. And you can see that now from
just a really simplified one group criticality relation. And if you raise the
fuel temperature, let's say the fuel gets
really hot because there's been some power
spike, you also want the reactor to shut itself down,
which you can see that it does. Let's make things
a little trickier. Let's now talk about
a sodium reactor. Fast reactor. This one relies a lot more
on fast fission of U238. So if we were to draw the two
cross-sections of 235 sigma fission and 238 sigma fission-- remember, uranium 235
looked like the one that we drew before, whereas
U238 goes something like that, with no actual scale given. I'm not going to even go there. But uranium 238 does not need
moderation for the neutrons to induce more fission. So let's now write the same
criticality reaction, which, again, is a super simplified
view of things, but that's OK. What would happen to each
of these terms in a sodium fast reactor if you
void the coolant? So nu won't change. What about sigma fission? Well, if the coolant goes
away, then on average there is fissile
materials contributing more to that cross-section,
but not that much. So if you want to
get technical, might be the slightest of increases,
but doesn't matter that much. What really matters, though,
is the stuff on the bottom. Sodium does have a low, but
non-negligible absorption cross-section. So if the sodium
were to boil away, then the absorption
would go down by a non-negligible amount. And then what about diffusion? Well, we've got the
formula for it up there. If there's not as much
coolant in the way, then the neutrons
are going to be able to get further on average. Let's say, they're not going
to be scattering around with as much of the sodium. So there might be a small
increase in diffusion and corresponding small
increase in buckling. But this is where the
one group kind of fails. What the sodium
is actually doing is providing a little
bit of moderation, so that some of those neutrons
when they bounce off of sodium leave the fast fission
region and get absorbed. And that's part of the
balance of the reactor. If all of the neutrons
are then born fast and don't really slow down
and just get absorbed, then you might have an overall
positive void coefficient. So this would tell you that
in a fast reactor where you're depending on your
coolant not just to cool the reactor, but
to absorb somewhat and to moderate
somewhat, you don't want to boil the coolant
in a fast reactor. And is a lot of the reason
why most fast reactor coolants tend to have
extremely high boiling points. Sodium is approximately
893 Celsius. Lead bismuth is
approximately 1,670 Celsius. Molten salt, about
1,400 Celsius. So all those coolant,
except for the sodium one, you'll melt the steel that
the reactor is made out of before your boil the coolant. So boiling the coolant is a
bad day in a fast reactor, because then things will
go from bad to worse, because in this case, the
feedback coefficient can be positive for
voiding the coolant. That's no good. So you want to keep
the reactor submerged. And that's another reason why
a lot of these fast reactors are what's called,
pool-type reactors. The reactor is not a
vessel with a bunch of piping under it that
can break and fail, but instead it's designed as
a huge pool of liquid sodium. And then the core is somewhere
in here with a bunch of pumps sending the coolant in and
back out, or through some heat exchange or something. So there's not really
any penetrations on the bottom up this pool. And you make sure
that you maintain, either when you have
sodium or lead bismuth eutectic, or liquid lead,
or some other fast reactor coolant. So these are some kind
of interesting scenarios to think about. I think one of them that
I put in the homework was imagine you have the
MIT reactor and replace the coolant with molten sodium. What's going to happen? Well, let's say you got
all the water out first and it wouldn't just blow up. What would actually happen
to the criticality relation? That's something I want
you to think about, because one of the big
problems on the homework is doing exactly
this for scenarios that have happened
to the MIT reactor, except for the sodium one. That's never happened
and hopefully never will. I can't even imagine. But now let's talk a
little bit about when you perturb a reactor by
doing something to it, putting the control rods
in, or pulling them out, or doing whatever you want. You're by definition
going to take one of our first assumptions
about how the neutron diffusion equation works and
throw it out the window. So we're now moving into
the transient regime. So to study what happens
in a reactor transient or when something changes
as a function of time, let's first go from k effective
to what we call k infinity. The multiplication factor
for an infinite medium. We're only doing
this because it's analytically easier
to understand and still gets the point across. So we'll say that our
k infinity is still a balance between
production and destruction. The difference is if we
have an infinite medium, there's no leakage. You can't leak out of an
infinitely sized reactor, should one ever exist. And so it just comes
out as nu sigma fission over sigma absorption. A much simpler form. And so now we can
write what would happen to the flux in the
reactor as a function of time. In this case, it's going
to be one over velocity. I'm going to make this a very
obvious wide v. That change in the reactor flux is going
to just be proportional to the imbalance now in the
number of neutrons produced and destroyed. So the number of
neutrons produced will be proportional
to very sharp nu sigma fission minus the number
of neutrons destroyed, sigma absorption, times
phi is a function of t. Y'all with me so far? So this right here
is a change, which is proportional to an
imbalance between production and destruction, times
the actual flux that you have in some given time. So to make this simpler,
let's multiply everything by v. Where's my green
substitute color? Multiply everything by v.
And the only unfortunate situation is we have a v
and a nu next to each other. I'm going to try to keep them
looking really different. Those go away. And then we end up with,
if we divide by phi, then those phi's go away. And we have phi prime over
phi, equals v nu sigma fission, minus v sigma absorption. And now we can start
to define things in terms of our
k infinity factor and a new quantity
I'd like to introduce called the prompt lifetime. It's a measure of how long a
given neutron tends to live before something happens to it. Before it's either absorbed
or leaks out, well, not from our infinite reactor. And so we can define this as
1 over their neutron velocity, times sigma absorption. And just to check
the units here-- velocities in meters per second. Macroscopic cross-sections
are in 1 over meters. So those cancel
out, and we're left with a total units of seconds. That's nice. We would want a mean
neutron lifetime, or a prompt lifetime to have
of seconds or time, at least. Yep. AUDIENCE: Can you say again
why the [INAUDIBLE] squared went away? MICHAEL SHORT: Why the d
phi dt squared went away? AUDIENCE: No, the
[INAUDIBLE] square. MICHAEL SHORT: Oh, OK. So that's because
we assume we're going to be analyzing
an infinite medium. So right here, this to
relabel these terms, this would be the
total production term. That right there
represents absorption. And that right there
represents leakage. But if we're analyzing
an infinite medium, you can't leak out, because it
takes up the entire universe and beyond, depending on what
you believe metaphysically. That's different costs. So this right here, we can
rewrite as 1 over lifetime. That makes it easier. And this right here,
if we note that nu-- that's a nu. I'm going to be really
explicit about that. Nu sigma fission over
sigma absorption. This kind of looks like this
is looking to be like k-- wrong color. --like our k infinity over lp. So all of a sudden we of
a much simpler relation. We have 5 prime over 5
equals k infinity minus 1 over the prompt
neutron lifetime. So if we solve this, this
is just an exponential. So we end up with our phi
as a function of t is-- whatever flux we started
at, like for your power in manipulations, it would be
whatever the neutron flux was before you touch the control
rod, times e to the t, or e to the that stuff, k
infinity minus 1 over lp, times t, which we can
rewrite as t over capital T. We're going to define
this symbol as what's called the reactor period. What the reactor
period actually says is how long before the flux
increases by a factor of e. And so this is actually
what that meter was measuring on the reactor. It's the reactor
period or the time it would then take for
the reactor's power to increase by a factor of e
because it's an exponential. To tell you what these
typical reactor periods tend to be for a thermal
reactor, t is about 0.1 seconds corresponding
to an average prompt neutron lifetime of 10 to
the minus 4 seconds. Seems fast, doesn't it? Like, really fast. So the question I asked you
guys is, why don't reactors just blow up? AUDIENCE: [INAUDIBLE]. MICHAEL SHORT: Yes,
there is something we've neglected from here. It's like what Sarah said. And it deserves its own board. There is a fraction
of delayed neutrons. We'll give that fraction
the symbol, beta. And for a uranium 235,
it equals about 0.0064. So there's less than a percent
of all the neutrons coming out of a reactor have some delay
to them, because they're not made directly from fission in
the 10 to the minus 14 seconds that we talked about
in the timeline. But they come out of
radioactive decay processes with delayed lifetimes
ranging from about 0.2 seconds to about 54 seconds. This is the whole reason why
reactors don't just blow up. So you can actually make a
reactor go super critical. But if the k effective
is less than 1 plus beta, then the reactor is not what
we call prompt super critical. And so the reason for that is,
let's say you raise the reactor power by some amount and the
k effective goes up to 1.005, there's still this fraction
0.0064 of the neutrons are not going to be
released immediately. They're going to be released not
in 10 to the minus 14 seconds, but in 10 to the 2 seconds. So a measly 15
orders of magnitude slower, meaning that
there's actually some ability for this reactor
to raise its power level. And these delayed neutrons,
even though that's such a small fraction,
takes the reactor period from its t infinity value
of about 0.1 seconds to about 100 seconds. So the same reactor
when you account for the delayed neutrons
increases in power by a factor of e. And it takes it about
100 seconds, which means this is totally controllable. Now I have a question
for you guys. Would you guys like me
to derive this formula, or do you want to go into
more of the intuitive implications of it? Because we can go either way. There is a formula
that will tell you what the reactor period and
time dependence will be. And you will hit it
in 22.05 probably. I can't guarantee it
because I'm not teaching it. Or we can talk a little bit
more about some of the intuition behind delayed neutrons. So a bit of choose
your own adventure. Math or intuition? AUDIENCE: Intuition. MICHAEL SHORT: Intuition. OK, that's fine. Good. So that was the derivation. I'll post that anyway,
if you guys want to see. I think in the Yip reading
it says, let's account for the delayed neutrons. Intuitively we find that
the answer ends up being-- so I'll skip the derivation. And it comes out to phi
naught e to the beta minus 1, times k minus 1 over lt, plus
beta phi naught over beta minus 1k, minus 1, times 1 minus e
to the beta minus 1k, minus 1 over l. OK, so left as an
exercise to the reader-- AUDIENCE: That's intuitive. MICHAEL SHORT: Yeah,
that's intuitive. But let's actually talk
about how intuitive it is. I do want to give you the
starting and the ending equation. And we will not go
through the rest. Yeah, Charlie? AUDIENCE: Should
we copy that down? MICHAEL SHORT:
No, you shouldn't. I'm going to scan
it for you guys. So don't bother copying it down. Let's talk about
where it comes from. And the answer may
astound you because we're going to bring right back the
idea of series radioactive decay. So let's say you want
to relate the change in the number in the
neutron flux to a 1 minus-- I'm going to take a quick
look at the original equation because I don't want
to screw that up. That's the first page, and
that's the one we want. Let's say we had some
equations that looked something like this. Phi plus phi naught times beta. This is the original
differential equation from whence it came. And the intuitive part
that I want you to note is that the jump from
changing k effective is moderated by this term
right here, 1 minus beta. So that's the fraction
of prompt neutrons, that as soon as you pull
the control rod out, that's your
instantaneous feedback. By instantaneous, I mean
on the order of, like, 10 to the minus 4
seconds, or something that you can't really control. This right here represents
the delayed fraction. This is as mathy as it's
going to get because you've chosen intuition. I think you have chosen wisely. It's going to be a more fun. So what this
represents right here is your kind of instant change,
because whatever you change k effective to, it's
going to be moderated by the prompt fraction, how
long the neutrons tend to take to undergo that feedback. Yes, Sara? AUDIENCE: Was that the average? MICHAEL SHORT: The average what? AUDIENCE: Average
neutron lifetime. MICHAEL SHORT: Yes, this is
the average neutron lifetime. So let's define the
average neutron lifetime as simply 1 minus beta times
the prompt neutron lifetime, plus the beta times some
delayed neutron lifetime. So what no book I've
ever seen actually says, this is what's referred to
as a Maxwell mixing model. It's just the simplest
thing to say, oh, if you want to get the
average of some variable, take the fraction of one
species times its variable, plus the fraction of the other
species, times its variable. Folks do the same thing
with electrical resistivity, thermal conductivity, or
any sort of other material property. And it is or isn't good
in some situations. Like, if you had a piece
of material made out of two different things-- let's say this had
thermal conductivity k1, and it had thermal
conductivity K2. Would a Maxwell mixing model
be appropriate to describe the flow of heat
across this thing? Probably not. But in the case
of neutrons where they're flying about like crazy
and their mean free path is much larger than the
distance between atoms, this works great. So we can define this
mean neutron lifetime and use that in this
equation right here. So this term right
here describes the instantaneous change. You pull the control
rods out, and fraction 1 minus beta neutrons
respond immediately. What about that
fraction of neutrons? Those are being produced
with a fraction beta depending on what
the flux was before, because they're still waiting to
decay from the old power level. Does anyone notice anything
suspiciously familiar about the final form of
this equation for flux? You've seen it
before with a couple of constants changed around. What about the form of
this differential equation? [INTERPOSING VOICES] It is exactly the same as
series radioactive decay. So the horrible derivation I
was going to do for you guys and we're not anymore is,
use an integrating factor. You solve it in
exactly the same way. You bring everything to
one side of the equation. Find some factor mu, that
makes this a product rule. Do a lot of algebra. And you end up with a very
suspiciously similar looking equation. So it's exactly the
same posing and solution as series radioactive decay,
with the difference being, that's the constant in
front of everything, instead of a bunch of
lambdas and fluxes. So what this says here is that
the flux as a function of time, this is the prompt
feedback right here, which says that
if-- let's graph it, since we're going intuitive. There's no room. Even those all boards are full. OK, here we go. If we graft time
and flux right here, what that part
right there says is that you're going
to get some sort of instantaneous
exponential feedback. But it's going to be moderated
by this one minus exponential on top. So you're going to end up with
a little bit of prompt feedback, this stuff right here. And then-- have to draw longer
because it's going to take forever-- you'll have some
delayed feedback, because you've got to
wait 100 or so seconds, or whatever that new
reactor period is, for the delayed
neutrons to take effect. And that's the whole reason you
could pull the control rods out at almost any speed you wanted
and the reactor doesn't just explode. If you pull the control
rods out fast enough, such that the change in k effective
is greater than beta, then the reactor goes
prompt super critical, which means you don't have any
delayed neutrons slowing down the feedback. And you've kind of turned
your reactor into a weapon. A very poor,
terrible weapon, but a prompt super critical
nuclear device, nonetheless. Did anybody pull out the
control rods too fast and the controls
took over for you? What about you guys in training? Did you ever do things when you
watched the automatic control take over? No? AUDIENCE: It'll just take over. AUDIENCE: Yeah, it'll kick you
off if you don't pay attention. MICHAEL SHORT:
That's what I mean. The machine takes over
and it will kick you off and stop responding to you. AUDIENCE: [INAUDIBLE]
horrible noise and so we don't want that. It's more to avoid
an annoying alarm. MICHAEL SHORT: I see. But the annoying
alarm is to stop you from doing something like that,
like, making the reactor go prompt super critical. AUDIENCE: [INAUDIBLE] MICHAEL SHORT: OK, so
that's what I would call the machine taking over. AUDIENCE: Oh, I see. AUDIENCE: It'll kick you on to
manual, and then [INAUDIBLE] still don't do
anything [INAUDIBLE].. MICHAEL SHORT: Yeah. So if your blood alcohol
level is above beta and you try and, let's
say, increase the reactor reactivity too much, it
will then take over, insert a control rod, make a horrible
noise, and say, go home, you're drunk. Something like that. OK, that makes sense to me. So what did your guys' reactor
power traces look like? Did they look
something like this, where there was an initial
rise as you pulled the control right out? And then after you pulled the
control rod out the power kept rising just a smidge, right? And what happened when you
put the control rod back in? Let's say you put the
control rod back in. You're going to get another
prompt drop, not equal to the same prompt
gain that you got, because now the reactor's
at a different flux, and then some asymptotic
feedback like that. And so this is why to
those who don't understand neutron physics, reactor
feedback is very non-intuitive. It's not a linear system. You can't just pull the control
right out and change the power accordingly. This is why there's automated
controls in systems to stop you in case, like I said,
if your blood alcohol content's above beta, which
is very low, by the way. Though you shouldn't
be drinking on the job, especially at a nuclear reactor. Plus, you're all under 21,
so what am I even saying? AUDIENCE: What is alcohol? MICHAEL SHORT: That's right. Good answer. What is alcohol? AUDIENCE: Is that going to
be covered in [INAUDIBLE] MICHAEL SHORT: That'll
be on the exam, yeah. AUDIENCE: What is alcohol? MICHAEL SHORT: Yeah, cool. So that's all I want to go
into for the intuitive stuff. And it's about 5 of 5 of. So I'd like to stop here and see
if you guys have any questions on neutron physics at a whole. Noting that we're going
to take Thursday's class and turn into a recitation. So I would like all of you guys
to look at the problem set, because it is posted. It is hard. Trust me. This one's a doozy. So I want to warn you
guys because you've got seven days to work on it. But I want you to look at it so
that we can start formulating strategies for the problems
together on Thursday, because there are
some tricks to it. You guys know me by now, right? There's always some
sort of a trick. Like, do you have to
integrate every energy to get the stopping power? No, you actually don't have
to do any integrals at all. But you can if you
want, and your answer will be more
accurate and correct. It'll just take
longer to get to. So there's a lot of diminishing
returns on these problems sets. If you're willing
to take an hour and think about
how can I do this simpler and with
fewer decimal points, you're probably onto something. And we'll work on those
strategies together. AUDIENCE: Is this due
next Monday as well? MICHAEL SHORT: Yes. So I posted it yesterday
at around noon or whatever the Stellar site says. I'll also teach you guys
explicitly how to use Janus. So we got a comment in from
the anonymous feedback saying, we have to use a
lot of software. Can we have some sort
of tutorial for dummies? Well, you guys aren't
dummies, but you still deserve a tutorial. So I will show you how
to export the data you'll need from this
problem set for Janus. So you can focus on the
intuition and the physics and not get frustrated with
getting data out of a computer. So any questions on anything
from the neutron diffusion equation? Yeah, Luke. AUDIENCE: I'm not real clear
on [INAUDIBLE] neutrons are and how those are different
from the prompt neutrons? MICHAEL SHORT: The
prompt neutrons come right out of fission. If we looked at
that timeline of, let's say the fission
event happens here. Two fission products
are released in about 10 to the minus 14 seconds. They move a little
further apart. And then some of them
just boil off neutrons, because they're so neutron
heavy, after around 10 to the minus 13 seconds or so. These right here
are prop neutrons, coming directly from the
immediate decay of neutron rich fission products. Some of the delayed neutrons
come from radioactive decay, but of the much later
fission products with much less
likely occurrences, which is why the
fraction is very low. But also, because it's
much longer half life, those delayed neutrons
take seconds, instead of pico seconds, to show up. And that's the whole basis
behind easier control and feedback a reactor. Good question. So anything starting
from neutron transport to simplifying to
neutron diffusion, to getting to this
criticality condition, making the two group
criticality condition if you want to have fast and thermal,
or any of the time dependent stuff that we intuited today. Yeah? AUDIENCE: So for that
cross-section [INAUDIBLE] you have there, so you have
one for 235 and one for 238. 235, it has to be thermal
neutrons [INAUDIBLE] fast? MICHAEL SHORT: Yep. AUDIENCE: And you said
that [INAUDIBLE] different [INAUDIBLE] as well it had-- if you have
[INAUDIBLE] into the-- MICHAEL SHORT: Was
it on the other board or from a different day? AUDIENCE: It was
a different board. MICHAEL SHORT: OK. AUDIENCE: Yeah, so could
you explain that graph? MICHAEL SHORT: Yes. So in this case-- let me get a finer chalk. This blue one would be
for low temperature, and this red one would
be for high temperature. So this blue graph,
there are resonances, which have very high values,
but they're very narrow. And because the width of a
resonance doesn't matter, it doesn't affect
the probability that a neutron scatters up
here and moves some distance down the energy spectrum. Thinner resonances tend
to get passed over, especially if your reactor's
full of hydrogen. Some of those neutrons will
be born and immediately jump into the
thermal region, where it's easy to tell how much
fission they'll undergo. As you go up in
temperature, you undergo what's called Doppler
broadening, which causes these resonances to
spread out and also go down in value. So the actual value
of the cross-section at these residences is lower,
but the widths are larger. So there's a higher probability
that a neutron scattering around and losing
energy will hit one of these higher
cross-section regions, called a resonance, at a
higher temperature. That's the difference there,
is these two plots show the same cross-section at
low and high temperature. These plots show the
difference between uranium 235 and uranium 238. Good question. Anyone else? Cool. OK, for the first
time in history, I'll let you out a minute early. Bring all your
questions on Thursday. So we'll start off
with a Janus tutorial. And then we'll start attacking
this problem set together.
