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visit MIT OpenCourseWare at ocw.mit.edu. MICHAEL SHORT: So I want
to do a quick review of what we did last time,
because I know I threw-- I think we threw the full six
boards of math and physics at you guys. We started off
trying to describe this general situation. If you have a small nucleus 1
firing at a large nucleus 2, something happens, and we
didn't specify what that was. A potentially
different nucleus 3 could come shooting
off at angle theta, and a potentially different
nucleus 4 goes off at a different angle phi. Just to warn you guys, before
you start copying everything from the board,
starting last week I've been taking pictures of
the board at the end of class. So if you prefer to look and
listen or just take a few notes rather than copy
everything else down, I'll be taking pictures of
the board at the end of class from now on and posting
them to the Stellar site. So up to you how
you want to do it. We started off with
just three equations. We conserve mass,
energy, and momentum. Mass and energy-- let's see-- come from the same equation. c squared plus T1 plus
M2 c squared plus T2 has to equal M3 c squared plus
T3 plus M4 c squared plus T4. We started off making
one quick assumption, that the nucleus 2, whatever
we're firing things at, has no kinetic energy. So we can just forget that. What we also said
is that we have to conserve x and y momentum. So if we say the x
momentum of particle 1 would be root 2 M1 T1
plus 0 for particle 2, because if particle 2 is not
moving, it has no momentum. Has to equal root 2
M3 T3 cosine theta, because it's the x
component of the momentum, plus root 2 M4 T4 cosine phi. And the last equation
for y momentum-- we'll call this x momentum,
call that mass and energy, call this y momentum-- was-- let's say there's no
y momentum at the beginning of this equation. So I'll just say 0 plus 0. Equals the y component of
particle 3's momentum, root 2 M3 T3 sine theta, minus-- almost did that
wrong-- because it's going in the opposite direction,
root 2 M4 T4 sine phi. We did something, and we
arrived at the Q equation. I'm trying to make sure we
get to something new today. So the Q equation
went something like-- and I want to make sure that
I don't miswrite it at all. So when we refer
to the Q equation, we're referring to this highly
generalized equation relating all of the quantities
that we see here. So I'm not going to go
through all of the steps from last time,
because, again, you have a picture of the
board from last time. But it went Q equals T1 times M1
over M4 minus 1 plus T3 times 1 plus M3 over M4 minus 2 root
M1 M3 T1 T3 cosine theta. And last time we talked about
which of these quantities are we likely to
know ahead of time and which ones might
we want to find out. Chances are we know all
of the masses involved in these particles,
because, well, you guys have been calculating that for the
last 2 and 1/2 weeks or so. So those would be
known quantities. We'd also know the Q value for
the reaction from conservation of mass and energy up there. And we'd probably be controlling
the energy of particle 1 as it comes in. Either we know--
if it's a neutron, we know what energy
it's born at. Or if it's coming
from an accelerator, we crank up the voltage on the
accelerator and control that. And that leaves us
with just three quant-- two quantities that
we don't know-- the kinetic energy of
particle 3 and the angle that it comes off at. So this was the highly,
highly generalized form. Recognize also that this is a
quadratic equation in root 3, or root T3. And we did something
else, and we arrived at root T3 equals
s plus or minus root s squared plus t, where s and t-- let's see. I believe s is root M1 M3 T1
cosine theta over M3 plus M4. And we'll make a
little bit more room. t should be-- damn
it, got to look. Let's see. I believe it's minus M4 Q plus-- oh, I'll just take a quick look. All right, I have
it open right here. I don't want to give you a
wrong minus sign or something. I did have a wrong minus sign. Good thing I looked. 1 times E1 over M3 M4. And so we started
looking at, well, what are the implications
of this solution right here? For exothermic reactions,
where Q is greater than 0, any energy E1 gets
this reaction to occur. And all that that says, well,
it doesn't really say much. All that it really
says is that E3-- I'm sorry-- T3-- and let me
make sure that I don't use any sneaky E's in there-- plus T4 has to be greater
than the incoming energy T1. That's the only real
implication here, is that some of the mass
from particles 1 and 2 turned into some kinetic
energy in particles 3 and 4. So that one's kind
of the simpler case. For the endothermic case,
where Q is less than 0, there's going to be some
threshold energy required to overcome in order to
get this reaction to occur. So where did we say? So, first of all, where
would we go about deciding what is the most favorable
set of conditions that would allow one
of these reactions to occur by manipulating
parameters in s and t? What's the first one that
you'd start to look at? Well, let's start by
picking the angle. Let's say if there
was a-- if we had what's called
forward scattering, then this cosine
of theta equals 1. And that probably gives
us the highest likelihood of a reaction happening, or
the most energy gone into, let's say, just moving
the center of mass and not the particles going
off in different directions. Let's see. Ah, so what it really
comes down to is a balance in making sure
that this term right here, well, it can't go negative. If it goes negative, then
the solution is imaginary and you don't have
anything going on. So what this implies
is that s squared plus t has to be greater or
equal than 0 in order for this to occur. Otherwise, you would
have, like I said, a complex solution to an energy. And energy is not
going to be complex. That means the
reaction won't occur. So this is where we
got to last time. Yes. AUDIENCE: When you say s
squared plus t is greater than 0 or greater than or equal
to 0, it's endothermic, wouldn't it also be
greater than or equal to 0 for an exothermic? MICHAEL SHORT: It would. But there's a
condition here that-- lets see. In this case, for
exothermic, Q is greater than 0 and that condition
is always satisfied. For an endothermic
reaction, Q is negative. So that's a good point. So if endothermic,
then Q is less than 0, and it's all about making
sure that that sum, s squared plus t,
is not negative. What that means is in
order to balance out the fact that you've
got a negative Q here, you have to increase T1 in order
to make that sum greater than or equal to 0. Yes. AUDIENCE: So that
condition, s squared plus t is greater
than or equal to 0, is that basically a condition
for the endothermic reaction to occur? MICHAEL SHORT: That's correct. If s squared plus t is
smaller than 0, which is to say that this whole sum
right here doesn't help you balance out the negative
Q, then the reaction is not going to happen. And something else might happen. So let's say you were looking at
a case of inelastic scattering where a neutron would
get absorbed by a nucleus and be re-emitted at a
different energy level. If the energy is too
small for that to occur, then the neutron is not
going to get absorbed. Instead it might bounce off
and undergo elastic scattering. And as a quick
flash forward, I'll show you a quick plot of elastic
and inelastic cross-sections that kind of hammers this home. You'll be looking at a
lot of these plots, that are going to be
logarithmic in energy space and probably logarithmic in
microscopic cross-section, to bring back that
variable from before. If you remember,
the cross-section is like the probability
that a certain reaction is going to occur. The larger the
cross-section, the higher the reaction rate for a
given flux of particles. And let's say we'll split
this into two things. We'll call it sigma elastic
and sigma inelastic. And we'll give them-- that will be-- oh,
we have colors. Let's just use those. Even better. Where's my second color? Under the paper. Awesome. So let's say the elastic
cross-section is in red, and the inelastic
cross-section is in green. And for white, we'll
plot sigma total. Usually, one of these
cross-sections for any old interaction-- I'm not even being
specific on which one. Let's just say neutrons
hitting something big-- would tend to look like this. There will be some insanity
here, that we'll discuss, and it might start to
increase a little bit as it goes to high energies. And this is definitely
not to scale. Just for the purposes
of illustration. The elastic cross-section
is going to look something like exactly this. See how closely I can
draw on top of it. So at low energies,
when a neutron can't be absorbed and re-emitted
at a different energy, the inelastic scattering
process can't happen. Which is to say that the
incoming kinetic energy-- so this log of E right
here, better known as T1 in the symbols up there-- it's not high enough to allow
inelastic scattering to occur. So if we want to graph
the inelastic scattering cross-section, it
will typically look like that, where once
you reach your threshold energy, determined by
that condition there, then the inelastic
scattering turns on and it's actually
able to proceed. So this is why we're getting
into these threshold reactions, because it helps
you understand why do some of the
cross-sections that we study have the shapes that they do. And this holds true
for pretty much every inelastic
cross-section I've seen, is they all--
almost all of them, if you're starting
at the ground state, require some initial
energy input to get going. Whereas elastic scattering
can happen at any energy. So let's write that last
condition right here. Elastic scattering,
which means things just bounce off like billiard
balls, you have Q equals 0. No energy changes
hands, so to speak. You just get some
kinetic energy from 1 being imparted to nucleus 2,
but you don't turn any mass into energy. So any questions here
before we go into-- yes. AUDIENCE: If theta isn't
0 for cosine theta, how would you plug it in? If you don't know what the angle
is, that it's not [INAUDIBLE]?? MICHAEL SHORT: So
the question is, if you don't know the angle,
what do you do about it, right? If you don't-- so in
this case, we've said, what is the bare minimum
threshold for this reaction to occur, and the best way for
that to happen is for theta to equal 0. If theta is larger,
that will actually mean that the reaction is not
allowed to proceed unless you get to an even higher energy. So this condition still holds. But if cosine-- so let's say
if cosine theta is less than 1, then the value of s goes down,
and that makes this condition harder to satisfy. So that's a good question. What that actually means is that
for certain nuclear reactions very close to the
threshold energy, only certain angles are allowed. I'm not going to get
into the nitty-gritty of which angles are allowed. I think it's-- I'll call it minutia for
the scope of this class, but it is in the
Yip reading, which I'll be posting pretty soon. But suffice to say that the
only time you can-- let's say if s squared plus t equals 0. The only time that can
happen is when theta equals 0 or cosine equals 1. And that means that the nucleus
can only recoil in a very, very narrow cone forward. As that energy increases,
the allowable angles start to increase further. Does that answer your question? AUDIENCE: Well, yes. So you're just
saying [INAUDIBLE] 0 is the minimum [INAUDIBLE]. MICHAEL SHORT: Or you'd say--
let's say if cosine equals 180. Then-- I'm sorry. If theta equals 180,
cosine would be negative 1. And that would be, let's say,
the least favorable condition. Yes. AUDIENCE: Why did you
put the sigma total under sigma elastic? MICHAEL SHORT: Oh, I'm saying--
so sigma elastic plus-- sorry-- sigma inelastic
would, let's say, give you the total
scattering cross-section. AUDIENCE: And then what
was the green line? MICHAEL SHORT: The green line--
oh, it is a little hard to see. The green line is the
inelastic cross-section. Yes. I can imagine from
back there the green and the white might look
a little similar, yes. OK, cool. Yes. Like one of these,
they give they give me an almost black one. That's about as
invisible as it gets. So make sure to
use visible colors. OK. So now let's take the case of
elastic neutron scattering. And can anyone tell me how
can we simplify the general Q equation for the
case of neutrons hitting some random nucleus? What can we start plugging
in for some of those values to make it simpler? What about the masses? What's M1 in atomic mass units? AUDIENCE: 1. MICHAEL SHORT: M1 is just 1 to
a pretty good approximation. It's actually 1.0087, which
we're going to say is 1. And what about M3? AUDIENCE: 1. MICHAEL SHORT: Is
also-- yes, also 1. If this is elastic
neutron scattering, the same neutron
goes in and goes out. So M1 and M3 are the same thing. What about M2 and M4? Have we specified
what this nucleus is? So what mass would we give it
if it's a general nucleus with N neutrons and Z protons? AUDIENCE: A. MICHAEL SHORT: A, sure. So A, that's again A
for the mass number. Cool. So with those in mind,
and then the last thing is we only have T1 and T3. Just for clarity, let's call
T1 Tin, like the neutron energy going in. And T3, we'll call Tout. So let's rewrite the
Q equation, the Q-eq, with these symbols in there. And the last thing
to note, what's Q for elastic scattering? 0, yes. Because no mass is turned
into energy or vice versa. So to rewrite the whole
Q equation, we'll get 0 equals Tin times M1 over
M4, which is just 1 over A, minus 1 plus Tout times
1 plus M3 over M4. M3 is also 1, M4 is also
A. And minus 2 root M1 M3. They're both 1. Tin Tout cosine theta. So this looks a
whole lot simpler. I'm going to do one quick
thing right here and take the minus sign that's hiding
in here outside this equation. It's going to make the
form a lot simpler. So I'm just multiplying
the inside and the outside of this term by negative
1, but hopefully you can see that it's
the same thing. It will just make the form
a lot nicer in the end. And so now we want
to start asking, what is the maximum
and minimum energy that the neutron can lose? So let's start
with the easy one. What is the minimum
amount of energy that the neutron could lose? Anyone? I hear some whispers. AUDIENCE: 0. MICHAEL SHORT: 0. And if the neutron comes in-- if theta equals
0, then you end up with actually Tin
will equal Tout. And, that way, let's say
delta T1 or delta T neutron could equal 0. So a neutron can lose at
least none of its energy in an elastic collision. Hopefully that makes
intuitive sense because we would
call that a miss. Now let's take the other case. At what angle would
you think the neutron would transfer as much energy as
possible to the recoil nucleus? So if we have a big
nucleus of mass A and we have a little neutron
firing at it, at which angle does it transfer
the most energy? Yes. AUDIENCE: Pi? If it's like-- MICHAEL SHORT: Exactly. AUDIENCE: [INAUDIBLE]. MICHAEL SHORT: At theta
equals pi, which means this-- we call this backscattering. So, yes, good one. I'll correct your
statement, though. You said if the
neutron just stopped and the nucleus moved forward. Does not happen in every case. For example, if you were to--
and I'd say don't try this at home, kids-- put on a nice helmet and
run charging at a truck, can you actually just stop cold? And we're not assuming any
bones breaking or anything. Chances are you'd
bounce right back off. Yes. That's the analogy I like to
give for what happens when a neutron scatters off uranium. It's like running at a
truck with a helmet on. It will just bounce back. So in the case of
theta equals pi-- so we're going to substitute
in theta equals pi. Therefore, cosine theta
equals negative 1, and we have an even
easier equation. 0 equals, let's just say,
Tout times 1 plus 1 over A. I'm going to arrange
these terms in order of their exponent for Tout
since that's our variable again. And if this stuff is negative
1, then the 2 minus signs cancel conveniently. And we have plus
2 root Tin Tout. Let's see. That's it. And we have minus Tin
times 1 over 1 minus A. Ideally, we'd like to try
to simplify this as much as possible. So let's combine. Let's try to get
everything in some sort of a common denominator,
because that would make things a lot easier. If we multiply each of
these 1's by A over A-- so let's put that as a step,
because we can totally do that-- we get 0 over Tout
times A over A plus 1 over A plus 2 root Tin
Tout minus Tin times A over A minus 1 over A. At this point, we can-- well, everything's in
common terms, right? We can just extend that fraction
sign and put the sign in here. Extend the fraction sign,
put the sign in here. And we'll just say that's
A. We'll say that's A. Last step that we'll do
is try and isolate Tout so at least one of our quadratic
factors is going to be simple, like 1. So next step,
divide by A plus 1. And then we get 0 equals
just Tout plus 2 over A plus one root Tin Tout minus Tin
times A minus 1 over A plus 1. Now we've got a simple-looking
quadratic equation, even though it's quadratic
in the square root of Tout. Yes. AUDIENCE: What happened to
the A from the denominator? MICHAEL SHORT: Let's see. AUDIENCE: Could
it be Tout over A? MICHAEL SHORT: What did I do? Did I miss an A
or dividing by A? AUDIENCE: The last
two equations. MICHAEL SHORT: It's
from back here? AUDIENCE: No, no, no. It's probably the
step you just did. MICHAEL SHORT: Just these steps. AUDIENCE: So you
divide by A plus 1. MICHAEL SHORT: Ah, I see. AUDIENCE: Should
it be Tout over A? [INTERPOSING VOICES] MICHAEL SHORT:
Yes, you're right. So I want to make
sure I didn't skip a step in dividing
an A. Let me just check something real quick. AUDIENCE: There
should've been an A in the minus 2 square root. MICHAEL SHORT: Oh, you're right. If we go back to
our Q equation-- let's see. There's an M4
missing, isn't there? That's it. Hah. See, this is what happens when
you don't look at your notes. I'll go back and correct
those, because then there should have been an over A.
There should have been an over A. There should
have been an over A. Thank you for pointing that out. And there should
have been another-- oh, in this case we can
just cancel all of the A's. I knew it came out
nice and clean. OK, cool. So at this point, this
is a quadratic in root Tout, where we have-- what are our a, b, and c terms
for this quadratic formula? So what's a first of all if
it's quadratic in root Tout? AUDIENCE: 1? MICHAEL SHORT: Just 1. That was part of the goal
of this manipulation, is to make at least one of
these things pretty simple. What's b? AUDIENCE: 2 over A plus
1 times radical Tin? MICHAEL SHORT: Yes. 2 root Tin over A plus 1. And c is just that
whole term right there. I'll do this up here. So then we can say root
Tout equals negative b plus or minus the
square root of b squared. So that's 4 Tin over
A plus 1 squared. Minus 4 times a times c,
so just minus 4 times c. So minus 4 times negative
Tin A minus 1 over A plus 1. So let's see what cancels. So, first of all, those
minus signs cancel. And everything has--
oh, and over 2a. Don't want to forget that. Over 2a, which is just 2. First thing we note is that
everything here has a 2 in it, either directly as a 2 or
hiding as a square root of 4. So we can cancel all of those. 4, 4, 4, 4. Let me make sure that minus
sign is nice and visible. What else is common
to everything here? Well, I'll tell you what. I'll write it all out
simpler without all the crossed-out stuff. Minus root Tin over A
plus 1 plus or minus root Tin over A plus 1 squared
plus Tin times A minus 1 over A plus 1. So with that written
a little simpler, what's also common and can be
factored out of everything? AUDIENCE: Square root of Ti? MICHAEL SHORT: That's right. Square root of Tin. Because there's a root Tin
here, and then you can-- everything's got a Tin
inside the square roots. You can pull that out. So we have a direct relation
between root Tout and root Tin. Minus 1 over A plus
1 plus or minus root 1 over A plus 1 squared
plus A minus 1 over A plus 1. What do we do here to
simplify all the junk in the square root? AUDIENCE: Multiply the right
side by A plus 1 over A plus 1. MICHAEL SHORT: That's right. You can always multiply by
something, better known as 1. And that gets
everything here-- just like there was a 2 or a root
4 everywhere in the equation, or there was a root
Tin and a root Tin everywhere else in
the equation, we'll do the same thing to get
the A plus 1 out of there. So we'll multiply this by
A plus 1 over A plus 1. I'll stick it over there. OK. And we get root Tout equals-- now everything has an A
plus 1, so let's bring all of those outside the fraction. Root Tin over A plus
1 times negative 1 plus or minus the square root
of 1 plus A minus 1, A plus 1. Starting to get a lot simpler. Let's see how much-- if I run
out of space for this one. So this stuff right
here is just A squared minus A plus A minus 1. The minus A and the
plus A cancel out. And then the plus 1 and
the minus 1 cancel out. And all that's inside the
square root is A squared. So the only hopefully
nonlinear board technique, I'm going to move to the left. And we end up with root Tout
equals root Tin over A plus 1. And all that's left
there is A minus 1 if we take the positive root. Almost done. Just square both sides. And we should arrive
at a result that might look familiar to some of you. Tout equals Tin times A minus
1 over A plus 1 squared. And we've gotten to
the point now where we can determine how much energy
the neutron can possibly lose or the recoil
nucleus can possibly gain in an elastic collision. It's this factor right here. I'll use the red since
it's more visible. This is usually referred to
in nuclear textbooks as alpha. It's sort of the maximum amount
of energy a neutron can lose or a recoil nucleus can gain. So what we've arrived at is
a pretty important result, that, let's say, the energy,
the kinetic energy of a neutron, has to be between its initial
kinetic energy and alpha times its initial kinetic energy. This right here is one
of the ways in which you choose a moderator
or a slowing down medium for neutrons in reactors. So it's this alpha
factor right here that really distinguishes
what we call a thermal-- or what is it? Like a light water reactor
or a thermal spectrum reactor from a fast
spectrum reactor. Let's look at a couple of
limiting cases to see why. Let's see. Anyone mind if I
hide this board here? Or you have a question? AUDIENCE: Yes. Can you explain why you ended
up dropping the negative case? MICHAEL SHORT: Let's see. If we took the negative case,
we'd end up with minus 1 minus A. You just have
an A plus 1 on the top. Yes. So in that case, you just have-- let's see. You just have root Tin, right? Let me see. AUDIENCE: Negative
root Tin actually. MICHAEL SHORT: Oh, yes. So that wouldn't make
very much sense, right? Yes. So in that case, well, you don't
want to have a negative energy. So that case doesn't
make physical sense. Thanks for making sure
we explained that. And did I see another question? Yes. AUDIENCE: Yes. What happened to
the coefficients you had before Tin? You had 4. You needed 4 or 2,
but [INAUDIBLE].. MICHAEL SHORT: Ah. OK. So what I did is I took the
square root of 4 out of every term inside the square root
and said, OK, they're all 2's. Just like in the next
step, I said, all right, there's all of these
A plus 1's, including all of these A plus 1's
squares inside the square root, and took that out. Or I think even over here. Yes, so the whole thing here
has been combine and destroy. Any other questions on what we
did here before I go on to some of the implications
of what we got? Cool. Let's look at a
couple limiting cases. I'll rewrite that inequality
right there because that's the important one of the day. So what is alpha for
typical materials? Let's say for hydrogen.
Alpha equals-- well, it's always A minus
1 over A plus 1 squared. And for hydrogen, A
equals equals 1, equals 1. And then we have 1 minus
1 in the numerator. Alpha equals 0. What this means is that
for the case of hydrogen, you can lose all of the neutron
energy in a single collision. That doesn't mean that you lose
all energy in every collision with a hydrogen atom
if you're a neutron, but it means that
you can lose up to all of your energy
in one single collision. And this is what makes
hydrogen such a good moderator or a slower down of neutrons,
is when it undergoes elastic scattering, especially
at energies below an MeV or so, which is where most of the
neutrons in the reactor are, it just bounces around. And the more it hits
hydrogens, the more it imparts energy to those
hydrogens and slows down. Why do we want to
slow the neutrons down in the first place? Well, that has to do with
another cross-section, that I'm going to draw
if I can find some chalk. Like I think I've
mentioned before, every nuclear reaction
has its own cross-section. And this time I'm going to
introduce a new one called sigma fission, the probability,
if a nucleus absorbs a neutron, that it undergoes
fission and creates more neutrons and like
200 MeV of recoil energy. So in this case, I'll
draw it for U235, since this is the one I pretty
much remember from memory. And it looks
something like that. So what you want is for the
neutrons to be at low energies. So this would be around the
thermal energy, better known as about 0.025 eV. Your goal is that the
more neutrons that you have in this energy region-- oh, that chalk
erases other chalk. The more neutrons you have in
that energy region, the higher probability you have a fission. So this is the basis
behind thermal reactors, is the neutrons all start here. They're born at
around 1 to 10 MeV. They don't undergo fission
very well at 1 to 10 MeV. So your goal as a
thermal reactor designer is to slow them down as
efficiently as possible. What's the most efficient
way to slow down neutrons? Cram the reactor
full of hydrogen. What's the cheapest and most
hydrogenous substance we know? Water. This is why water makes such
a good reactor moderator. It's pretty cool. There's also lots of other
reasons that we use water. It's everywhere, which is
another way of saying cheap. It's pretty chemically inert. There are corrosion
problems in reactors, but it doesn't just
spontaneously combust when you see air, like sodium
does, another reactor coolant. It takes a lot of
energy to heat it up. So it's specific heat
capacity, the CP of water, if you remember-- I think it's-- was it
4.184 joules per gram? That's a point. One of the highest
substances that we know of. So you can put a lot
of that recoil energy or a lot of heat
energy into this water without raising its
temperature as much as a comparative substance. Metals can have heat capacity's
like three or four times lower. So you wouldn't necessarily
want to use a metal coolant. Or would you? In what cases would you
want to use a metal coolant for a reactor? Has anyone ever heard of
liquid metal reactors before? Just a couple. Good. I get to be the first
one to tell you. I did my whole PhD on alloys
for the liquid lead reactor. So let's take a look at lead,
which has an A of about-- let's call it 200. I think there's some
isotopes, like 203 or so. There's probably an
isotope called lead 200. What would alpha be for lead? Well, let's just
plug in the numbers. A minus 1 is 199. A plus 1 is 201. Square that. Almost 1. Almost. This means that when neutrons
hit something like lead, basically don't slow down. They can lose at least
none and at most almost none of their energy. And this is the
basis behind what's called fast reactors if you
want to use a coolant that keeps the neutrons very fast. Because for uranium 238,
there's what's called a-- well, what you do is you
want to capture neutrons with uranium 238, make plutonium
239, and then breed that. Or uranium 238 has got its
fast fission cross-section. I don't think I want to get
into bringing it on the screen today since we're
almost five of five of. What I will say is there's
lots of other reactor coolants besides water. And it sounds to me like
almost no one had heard of a liquid metal reactor. Why would you want
to use a liquid metal as a coolant besides keeping
the neutrons at high energy? Anyone have any ideas? What sort of properties do
you want out of a coolant? Not even for a reactor
but for anything. AUDIENCE: Heat transfer. MICHAEL SHORT:
Good heat transfer. Metals are extremely
thermally conductive. So if you want to get the
heat out of the fuel rods and into the coolant and
then out to make steam for a turbine, liquid metals
are a pretty awesome coolant to use because they conduct
heat extremely well. What else? Let's try and think now. If you were a reactor
designer, you don't just have to make the reactor
work but you want to make it avoid accidents. What sort of thermodynamic
properties about metals could prevent accidents
from happening? AUDIENCE: They solidify. MICHAEL SHORT: Yes. So there's one problem. They could solidify. So coolants that have
been chosen for reactors have been things like
sodium, which melts just below 100 Celsius, liquid
lead-bismuth, which melts at 123 Celsius. And I know because I've
it in a frying pan before. So I did four years of research
on liquid lead-bismuth, and if there's anyone that's
gotten enough exposure to that, it's me. It does not seem to
have affected my brain too much because I only
made like two major mistakes on today's board. Good enough. Yes. So we've got hundreds of pounds
of lead-bismuth sitting around. It's pretty inert stuff. It's really dense, so it
can store a lot of heat. The other thing is
the boiling point. Anyone know what
temperature metals boil at? [INTERPOSING VOICES] MICHAEL SHORT:
Extremely high, yes. Sodium boils at approximately
exactly 893 degrees Celsius. Liquid lead-bismuth boils at
approximately exactly 1,670 Celsius. You'll actually melt the
steel that the reactor is made of before you
boil off your coolant. And if you boil
your coolant, you have no way of
cooling the reactor, and that's something
you want to avoid. Water boils at
approximately 325-- let's say 288 to 340 Celsius
depending on the pressure that's used in the reactors. And that does get reached
sometimes, especially in accident conditions. So if you want to make something
relatively Fukushima-proof, then you don't want
the coolant to boil. So use a liquid metal, which
introduces other problems. It also introduces
some other problems. I'm going to flash forward a bit
to neutrons and reactor design. Because it does take time for
this scattering to happen. These collisions, they
happen pretty quickly but they do take a
finite amount of time. And in the meantime,
you can have what's called
feedback coefficients, natural bits of physics that
help your reactor stay stable or they don't, depending
on whether it's called negative or positive feedback. So we can have either
negative or positive feedback. I'll give you one simple
example that I'll just introduce conceptually,
and we'll actually explain it a little
mathematically later in the course. Let's talk about
coolant density. If you were to heat
up your reactor and the coolant were
to get less dense, what do you think would
happen to the reaction rate of, well, anything--
scattering, fission, absorption? AUDIENCE: Go down. MICHAEL SHORT: It
should go down. Why do you think that is? AUDIENCE: Because there's
not as many particles that's close together, so [INAUDIBLE]. MICHAEL SHORT: Exactly. Yes. To reintroduce a bit of the
cross-section stuff I mentioned last time, the
microscopic cross-section is the probability that,
let's say, one nucleus hits one other nucleus. If you then multiply
by the number density or how many of
them are there, you end up with the
macroscopic cross-section. So I'll label this as
micro, label this as macro. And then the macroscopic
cross-section times your neutron flux
gives you your reaction rate. So if you want to get less
moderating happening, or less fission, or less
absorption, the simplest way is-- well, that's a
property of the material. That's whatever your
reactor is doing. You can decrease
the number density by heating things up and
decreasing the density. So this is one of those cases
where you can use the reactor to quickly respond with physics
before you could respond with human intervention. If you want, let's
say, a extra power transient or a sudden
increase in heat to slow down the nuclear
reaction and not speed it up, you'd pick a moderator
that behaves in this way. So in this case, water
would get less dense, it would moderate less
well, and put fewer neutrons in the high-probability
fission region. Then let's think
about what happens if you're depending on
your neutrons to stay fast or at high energy. Let's say you were to
have a really bad day and boil your liquid sodium. All of a sudden, what
little moderating power exists in that sodium
disappears or gets even lower, and that would cause your
reactor power to increase. So one of the dangers of
fast spectrum reactors is positive-- what's called
positive void coefficient. Or if you make a bubble
of gaseous sodium, your power increases
rather than decreases. That would increase the heat. That would cause
the power to go up. That would increase the heat. That would cause
the power to go up. Luckily, there are many,
many other negative feedback mechanisms that
could be built in to make sure the overall
feedback coefficients are still negative. This also lets us understand
a little bit about what went wrong at Chernobyl. And I'll give you a
1-minute flashforward. Because the control
rods that were made of-- let's see. I don't remember what the
composition of the control rods is, but they were
neutron absorbers. The control rods in
Chernobyl looked something like this, where there
was the absorber here. And this was-- they
were graphite tipped. And as you lower that graphite
down into the reactor, you're all of a sudden
introducing something. Well, that's an OK moderator. For carbon, let's
say A equals 12. So our alpha will be A
minus 1 over A plus 1. I'm not going to write
almost equal to 1 because that isn't
quite almost equal to 1. What does this actually equal? Let's see. Let's just say
definitely less than 1. There is some moderating
power to graphite. It's also a very bad absorber. And what this meant, that
as you lowered those control rods into the reactor,
you suddenly introduced a little more moderation when
things were already going bad, and that caused the power
level to increase further. There were other
problems, like it was designed so that if you
boiled some of the coolant, you would have
positive feedback. And that is the sort of
1-minute synopsis as to what all went haywire at Chernobyl. But we'll be doing a second
by second or, in some cases, millisecond by
millisecond play of what went wrong with Chernobyl. And we could probably do
the same for Fukushima now, now that we understand what
happened, based on the physics you'll be learning
in this course. And this is actually a
perfect stopping point, because next up we're
going to be looking at the different processes
of radioactive decay, many of which are a
simplification of this Q equation, and I
think some of which are probably
familiar to you guys, because radiation decay is
part of the normal lexicon, especially nowadays. So since it's five of
five of, do you guys have any questions on
what we've covered so far? Yes. AUDIENCE: [INAUDIBLE] if you
have water as your coolant and it gets too
hot, [INAUDIBLE].. MICHAEL SHORT: Yes. AUDIENCE: Right. And that will decrease
the reaction rate. MICHAEL SHORT: Um-hm. AUDIENCE: And that's how
like the [INAUDIBLE].. MICHAEL SHORT: It's actually-- are you asking
about it the water feedback is part of the backup? That's your primary
line of defense. Your backup is
human intervention, because, compared
to physics, humans are really, really slow,
like many orders of magnitude slower. It takes microseconds for
things to thermally expand. It definitely takes more than
seconds for a human to respond. Anyone ever done those tests
where you have a light blinking and you have to hit a button
the second you see the light? What's the fastest any of
you guys have ever responded? Anyone remember? Anyone beat a second? AUDIENCE: Maybe. MICHAEL SHORT: Maybe. A tenth of a second? [INTERPOSING VOICES] MICHAEL SHORT: And there all
you have to do is hit a button. All you have to do is
hit the only button when you see the only light. What if you're
piloting something that's about as complicated
as the space shuttle but more likely to explode? What do you think your
reaction time will be? Probably long. You'll probably have
to pull out the manual. And probably you'll have
to RTFM for a little while. And maybe you'll find out what
you have to do and maybe you won't. So it's actually
operator error that has caused most of the near
misses or actual misses in nuclear reactors. The physics, except for
the Russian RBMK design that was for
Chernobyl, usually it's human error that's the
downfall of these things. So by understanding
the physics here, we can rely on it
to keep things safe. Yes. AUDIENCE: When you
have a lead reactor-- or I don't know [INAUDIBLE] one
of these, but what is like-- how do you cool the lead
once it starts getting hot? MICHAEL SHORT:
Ah, good question. How do you cool the liquid lead? You can't send liquid led
through a turbine, right? So at some point you've got
to make steam and use that to drive a turbine. You can use what's
called a heat exchanger. At its simplest, you can think
of it like a couple of tubes where the lead is
going through here and the steam is
going through here. And they have a very thin
barrier between them, so you have all this heat moving
from the lead, which is hotter, to the steam, which is colder. They actually have built a
bunch of these led reactors. AUDIENCE: Is that real? MICHAEL SHORT: Yes. The Russian fast attack
subs, the alpha class subs, were powered by and are
powered by liquid led reactors. They're the only reactor
that can outrun a torpedo. So when you have a liquid
lead reactor powering and you've got a panic
button that says, forget the safety systems. Outrun a torpedo. You have a choice between maybe
dying in a reactor explosion and definitely getting shot out
of the water with a torpedo. You do whatever you can. And these subs only run
two or three not slower than a torpedo. So just like that
old algebra problem, if this guy leaves
Pittsburgh at 8 AM traveling 40 miles an hour, and I'm
trying to get to Boston, 30 miles an hour, if a
torpedo leaves one sub moving this velocity and the
alpha attack sub senses it from this distance
and starts moving at a similar velocity,
chances are the torpedo runs out of steam before
it reaches the alpha sub. And that's only because
they can have an extremely compact liquid lead nuclear
reactor at the power source. AUDIENCE: So can
you [INAUDIBLE]?? How do you move [INAUDIBLE]? MICHAEL SHORT: Good question. How do you use the-- how do you move the liquid lead? You can move it by
natural convection but that's extremely slow. So there are multiple
ways of moving it. One of the cool ones is called
an EM or Electromagnetic pump. It induces eddy currents
in the liquid lead because it's also a conductor. And those eddy currents couple
with the EM field from the EM pump and cause the lead to
just start moving on its own. So it's a no moving parts pump. The only problem is it's
like 1% or 2% efficient. Yes. So they only use
those on the subs, but you can use EM pumps to
move conductive coolants. So I think it's pretty awesome. And there have there have
been land submarines. In fact, there's a company
called AKME Engineering in Russia that's
trying to commercialize a small modular
liquid lead reactor. The other nice thing about
these liquid metal coolants is you can make
the reactors much, much smaller and denser than
in a light water reactor. In a light water reactor,
you're relying on a lot of water to cool things
and a lot of water to be there to
moderate your neutrons. In a liquid metal reactor,
where you don't need moderation, well, you don't
need-- all you need is enough coolant
to keep things cool. So you can tighten stuff up
and make it more compact. So that's one of the nuclear
startups coming out nowadays. This is an awesome
time to be in nuclear. When I started nuclear, there
were approximately exactly zero nuclear startups. Like TerraPower
didn't even exist yet. Now there's something
like 52 in the US and others around the world. So like this is the time
to be in nuclear if you're up for startups and not
just working in academia, or a lab, or a big corporation. There's a lot of little
companies now doing some crazy things based on
some pretty good physics. So maybe time for one more
question before I let you guys go. If not, then I'll see
you guys on Tuesday when we start radioactive decay.
