# Neutron to start the chain reaction

sstone
Hello,

I can't find where do we get the starting neutron(s) for the chain reaction in the reactor. I'm aware that then the splitted atom emits enough(in fact even too much?) neutrons to selfsustain the reaction, which are then slowed down to be able to split uranium and so on.

So where do one get the triggering neutron, to start the chain reaction?

Thanks.

Azael
There is a small probability that the Uranium nucleus can spotantiously fission, this releases neutrons. So uranium triggers its own chain reaction.

In nuclear weapons there is a source of neutrons included to trigger the reaction, something like a beryllium and polonium mix. When beryllium gets hit by alpha particles from the polonium it emits neutrons. This to ensure that there is indeed enough neutrons around exactly when needed.

Staff Emeritus
Hello,

I can't find where do we get the starting neutron(s) for the chain reaction in the reactor. I'm aware that then the splitted atom emits enough(in fact even too much?) neutrons to selfsustain the reaction, which are then slowed down to be able to split uranium and so on.

So where do one get the triggering neutron, to start the chain reaction?

Nuclear reactors contain special 'startup' sources to provide sufficient neutron activity, so that the activity in the core can be monitored. Otherwise, one would approach criticality without sufficient knowledge of knowing how close one is. The objective is to maintain control of the reactor.

While the (alpha, Be) sources are possible, e.g. Po-Be, Ra-Be or Pu-Be, they are not typically used. Rather the startup sources use Sb124-Be, which is a photoneutron source in which a high energy gamma photon knocks out a neutron from Be-9, which becomes Be-8, which promptly decomposes to 2 alpha particles. Sb-124 (which is formed from neutron absorption (activation) by Sb-123) emits a 1.6 MeV gamma.

See pages 37 & 38 of this document - http://lrs.web.psi.ch/educational/courses/2006_EPFL_DOCTORAL_PSI_COURSE/week3/Week_3_Lecture_5.pdf

Once the reactor has completed a few cycles of operation, it might have sufficient transuranics in the twice burned fuel to enable a 'sourceless' startup. Otherwise, the source stay in the reactor where they are reactivated by the neutron flux.

Azael
Nuclear reactors contain special 'startup' sources to provide sufficient neutron activity, so that the activity in the core can be monitored. Otherwise, one would approach criticality without sufficient knowledge of knowing how close one is. The objective is to maintain control of the reactor.

While the (alpha, Be) sources are possible, e.g. Po-Be, Ra-Be or Pu-Be, they are not typically used. Rather the startup sources use Sb124-Be, which is a photoneutron source in which a high energy gamma photon knocks out a neutron from Be-9, which becomes Be-8, which promptly decomposes to 2 alpha particles. Sb-124 (which is formed from neutron absorption (activation) by Sb-123) emits a 1.6 MeV gamma.

See pages 37 & 38 of this document - http://lrs.web.psi.ch/educational/courses/2006_EPFL_DOCTORAL_PSI_COURSE/week3/Week_3_Lecture_5.pdf

Once the reactor has completed a few cycles of operation, it might have sufficient transuranics in the twice burned fuel to enable a 'sourceless' startup. Otherwise, the source stay in the reactor where they are reactivated by the neutron flux.

Thanks for that information. It never occurred to me that a reactor needs a startup source aswell.

Dearly Missed
Thanks for that information. It never occurred to me that a reactor needs a startup source aswell.
Azael,

Actually after the reactor has been run; you don't even "need" the startup source, since
there will be enough radionuclides present as a result of fission that can provide the
initial neutrons.

However, as a matter of a condition on the operating license; the operator is required to
have a startup source in the core when it is sub-critical in any case. The instrumentation
for monitoring the core and its criticality are based on detecting neutrons.

Although with a fresh core, there are probably enough stray neutrons around to start the
reaction; you can't count on that. So the operator is required to have a source in the
reactor when it is subcritical, as Astronuc points out; so that one can monitor the
condition of the core.

Dr. Gregory Greenman
Physicist

sstone
Thank you all for the info.

curie
Further to the point that a extraneous neutron source is not always required, consider the natural fission reactor that is believed to have existed at Oklo, Africa a couple of billion years ago.

Staff Emeritus
Further to the point that a extraneous neutron source is not always required, consider the natural fission reactor that is believed to have existed at Oklo, Africa a couple of billion years ago.
This is true. It depends on kinf and size, which in part determines keff. Composition and homogeneity are also factors.

Dearly Missed
Further to the point that a extraneous neutron source is not always required, consider the natural fission reactor that is believed to have existed at Oklo, Africa a couple of billion years ago.
curie,

Without the neutron source; the Oklo could have just sat there in a critical state; but with
no neutrons in it. However, given that there are extraneous neutrons; it is only a matter
of time before one finds its way to the reactor, and we get a self-sustaining chain reaction.

But there DOES need to be an initial neutron - provided by cosmic rays, radioactive decay...
whatever.

In a manufactured reactor, that has been operating for even a short period of time; the
neutrons to restart the reactor can come from the fission products; some of which are
neutron emitters.

For a brand-new manufactured reactor; then you need a source. You don't want to be
trying to take the reactor critical and not have any neutrons in it. The instrumentation
as to the criticality of the reactor responds to neutrons. You can put the reactor into
a critical state - or even a super-critical state - without neutrons; and without those
neutrons, you wouldn't know that you had reached the critical or super-critical state.

That's why reactor operators are REQUIRED to have a source in the reactor at low
power and sub-critical conditions.

Dr. Gregory Greenman
Physicist

Emfuser
ou can put the reactor into
a critical state - or even a super-critical state - without neutrons; and without those
neutrons, you wouldn't know that you had reached the critical or super-critical state.

What the...?

That would be akin to suggesting that I could start-up, run, and accelerate an internal combustion engine without fuel.

No neutrons, no criticality. No criticality, no super-criticality.

That's why reactor operators are REQUIRED to have a source in the reactor at low
power and sub-critical conditions.

Dr. Gregory Greenman
Physicist

I'm not sure who you're generalizing this to, but operating power reactors that aren't brand new most certainly do not have this requirement. There is more than enough neutron activity coming from the spent fuel going through multiple cycles for the nuclear instrumentation to pick up.

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theCandyman
What the...?

That would be akin to suggesting that I could start-up, run, and accelerate an internal combustion engine without fuel.

No neutrons, no criticality. No criticality, no super-criticality.

I think Mobius is talking about the reactor being critical as in its size is big enough to sustain a reaction if there are neutrons. For example, 10 kg of Pu is supercritical, but not much will happen until there are incident neutrons.

Dearly Missed
What the...?

That would be akin to suggesting that I could start-up, run, and accelerate an internal combustion engine without fuel.

No neutrons, no criticality. No criticality, no super-criticality.
Emfuser,

Then you don't understand the concept of "criticality".

The car is NOT a critical system - the analogy doesn't hold.

Criticality is a property of the geometry and the materials; NOT the neutrons.

Theoretically, you could have a system that is "critical" even when neutrons
are not present.

[ A critical system will be "steady-state" only if the neutron distribution that is
present is the fundamental eigenfunction of the critical system. ]

You can assemble a system that is super-critical; even without neutrons.

That's why nuclear weapons have "initiators":

http://en.wikipedia.org/wiki/Nuclear_weapon_design#Neutron_trigger_.2F_initiator

The nuclear weapon has a device to put neutrons into the system; in case there aren't
any there to start the reaction when the device assembles. You wouldn't call an
assembled nuclear weapon "sub-critical" would you?

Of course not! You can assemble a system that is critical, or even super-critical; but
nothing will happen if there are no neutrons present.

That's because the concept of "criticality" is NOT a property of the neutron population;
it is a property of the materials and the geometry that they are in. Criticality tells you
what the system would do if neutrons are present; whether or not those neutrons are
actually present.

The word "critical" comes from mathematics. If the eigenvalue of the transport equation
for the system is exactly unity; mathematically the system is "critical". Note that the
solution of the transport equation is only dependent on the materials and geometry;
NOT the neutron population. In fact, it CAN'T be - because the eigenvalue system is
singular - being both linear and homegeneous - any constant multiple of a solution is also
a solution; including a multiple of zero. The criticality doesn't depend on the neutron
population.

The "reactivity" on the other hand, DOES depend on the neutron population. A system
could be exactly critical; but with a neutron population that isn't the fundamental
eigenfunction. Such a system would not be steady-state; the reactivity would be a
finite non-zero value until the system evolved into one where the neutron population
is the fundamental eigenfunction.

It's really the matematical system that is "critical"; and that just depends on materials
and geometry, and NOT the neutrons.

Dr. Gregory Greenman
Physicist

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Emfuser
Good lord... why is it always the book-bound guys that come up with this stuff? If you want to talk in the purely theoretical realm, then by all means, feel free to talk right out of a reactor dynamics book all day. However if you want to talk about the world of operating power reactors (which you did), then I suggest you step away from the chalkboard and try not using the most mathematically bound definition of criticality that you can find in place of the concept of a critical mass or critical geometry.

I passed all my reactor theory & neutron transport classes without problem. I understand full well what you're talking about, and then some. However I, just like every other nuclear professional who isn't holed up in a university, office, or other paper-bound world, recognize that the zero solution is always immediately discarded as trivial and that nobody ever talks about criticality exclusively in terms of critical geometry. It's always in the context of neutron population and 'k'.

That's why this statement:
Morbius said:
That's why reactor operators are REQUIRED to have a source in the reactor at low
power and sub-critical conditions.
... is false. The zero solution to the transport equation means nothing to us.

I'm sorry if this seems a bit curt, but in the world of operating power reactors, or any related work or study function, critical is not taken in the literal (but wondefully and uselessly abstract) mathematical definition. In the working nuclear world, critical has EVERYTHING to do with your neutron population and how it behaves. The reactor is real, the mathematical model is not. One is reality, the other abstractly attempts to model it.

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quetzalcoatl9
Good lord... why is it always the book-bound guys that come up with this stuff? If you want to talk in the purely theoretical realm, then by all means, feel free to talk right out of a reactor dynamics book all day. However if you want to talk about the world of operating power reactors (which you did), then I suggest you step away from the chalkboard and try not using the most mathematically bound definition of criticality that you can find in place of the concept of a critical mass or critical geometry.

I passed all my reactor theory & neutron transport classes without problem. I understand full well what you're talking about, and then some. However I, just like every other nuclear professional who isn't holed up in a university, office, or other paper-bound world, recognize that the zero solution is always immediately discarded as trivial and that nobody ever talks about criticality exclusively in terms of critical geometry. It's always in the context of neutron population and 'k'.

That's why this statement:

... is false. The zero solution to the transport equation means nothing to us.

I'm sorry if this seems a bit curt, but in the world of operating power reactors, or any related work or study function, critical is not taken in the literal (but wondefully and uselessly abstract) mathematical definition. In the working nuclear world, critical has EVERYTHING to do with your neutron population and how it behaves. The reactor is real, the mathematical model is not. One is reality, the other abstractly attempts to model it.

if i may: LOL...are you serious? is speaking about nuclear physics in mathematical terms silly?

Homer Simpson works at a power plant too...i think that you are way over your head in debating Morbius on these kinds of issues - why don't you search the PF archives...

Dearly Missed
Good lord... why is it always the book-bound guys that come up with this stuff? If you want to talk in the purely theoretical realm, then by all means, feel free to talk right out of a reactor dynamics book all day. However if you want to talk about the world of operating power reactors (which you did), then I suggest you step away from the chalkboard and try not using the most mathematically bound definition of criticality that you can find in place of the concept of a critical mass or critical geometry.\
Emfuser,

The use of the word "critical" comes from the mathematics. It is YOU that is using it
in a non-standard fashion

I passed all my reactor theory & neutron transport classes without problem.

You may have passed; but this is one part of transport theory that you have
WRONG!!!

I understand full well what you're talking about, and then some.

Asserting facts not in evidence. I see no evidence of "..and then some".

However I, just like every other nuclear professional who isn't holed up in a university, office, or other paper-bound world, recognize that the zero solution is always immediately discarded as trivial and that nobody ever talks about criticality exclusively in terms of critical geometry. It's always in the context of neutron population and 'k'.

Yes - 'k"!! What is "k". It is an eigenvalue of the transport equation. [There are
others.] The existence and value of "k" exists INDEPENDENT of the value of the eigenfunction;
whether that eigenfunction is the trivial zero, or a non-zero eigenfunction.

Look at any expression for "k". Do you see anything that has to do with neutrons like a
flux? No - you will see only material and geometric properties.

That's why this statement:

... is false. The zero solution to the transport equation means nothing to us.

I'm sorry if this seems a bit curt, but in the world of operating power reactors, or any related work or study function, critical is not taken in the literal (but wondefully and uselessly abstract) mathematical definition. In the working nuclear world, critical has EVERYTHING to do with your neutron population and how it behaves. The reactor is real, the mathematical model is not. One is reality, the other abstractly attempts to model it.

The above shows that you really DON'T understand some very important features about
the operation of a nuclear reactor. I would defy you to calculate reactor transients,
and do reactor "noise" analysis with your concept of what criticality is.

It is actually VERY IMPORTANT to the transport computer codes that are used in the
design and safety analysis of the reactor. You wouldn't know about that - because all
that work and analyis is done BEFORE you get your hands on the reactor.

Your concept of "reality" being divorced from the mathematics; what I call the
"seat of the pants" reality is really quite limiting. Without the mathematics, which you
seem to disdain; the reactors wouldn't work as well or as safe as they do.

The term "criticality" comes from the mathematics. The concept of "k" is an
eigenvalue; which comes from the mathematics.

It is the mathematics that keeps us well grounded in the physics; NOT the "seat of the
pants" understanding.

You might find it illuminating to consider the problem I posed in my previous post.
Let's consider the detonation of "Little Boy". The fissile material in Little Boy has
assembled, but the initiators haven't fired yet. Let's suppose additionally that there
were no stray neutrons around.

Little Boy has assembled its mass of U-235; but for an extremely brief period of time,
there are no neutrons present, as of yet. Consider:

Is Little Boy critical?

What is "k" of Little Boy? Does it exist? If so; can you say something about its value?

Or consider the following. I have an EXACTLY critical reactor with the fundamental
mode flux therein. I send a pulse of neutrons into the reactor.

Have I changed the criticality of the reactor by adding the neutron pulse?

Have I changed "k" for the reactor by adding the neutron pulse?

Have I changed the reactivity of the reactor by adding the neutron pulse?

Dr. Gregory Greenman
Physicist

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Dearly Missed
Homer Simpson works at a power plant too...i think that you are way over your head in debating Morbius on these kinds of issues - why don't you search the PF archives...
quetzalcoatl9,

Yes - it was nuclear power plant operators like "Homer Simpson" working
at Three Mile Island Unit 2 that needlessly plunged the nuclear industry
into a tail-spin for more than a quarter century; because they didn't
understand the physics of a reactor as well as they thought they did.

That was the subject of a pevious discussion here, not long ago.

It's like the old adage from the aerospace industry; "Planes are designed
by people with PhD degrees, built by people with Master degrees..."

There's a REASON they have the PhDs design the planes.

Dr. Gregory Greenman
Physicist

curie
curie,
Without the neutron source; the Oklo could have just sat there in a critical state; but with
no neutrons in it. However, given that there are extraneous neutrons; it is only a matter
of time before one finds its way to the reactor, and we get a self-sustaining chain reaction.

But there DOES need to be an initial neutron - provided by cosmic rays, radioactive decay...

I was going to say I meant source in the "conventional sense" of a neutron source, but that is actually incorrect & lazy. I still don't think the initial neutron would have to literally be a neutron external to the assembly. Could it have been the initial neutron was from a spontaneous fission (SF) inside the assembly that happened to "get lucky" at a time when the reflecting & moderating properties of the assembly were also just right? If so, then your inclusion of radioactive decay as a source of initial neutron could be said to include SF.

Btw, I would be interested to read your input on the questions you posed on the example scenarios.

Dearly Missed
Btw, I would be interested to read your input on the questions you posed on the example scenarios.
curie,

I'll give Emfuser a chance to answer.

If not - then I will give a full explanation to each of those questions.

Dr. Gregory Greenman
Physicist

Emfuser
if i may: LOL...are you serious? is speaking about nuclear physics in mathematical terms silly?

Homer Simpson works at a power plant too...i think that you are way over your head in debating Morbius on these kinds of issues - why don't you search the PF archives...

Not really

Morbius is the typical paper-bound scientist. Completely stuck in a paper world with a very narrow, rigid view of things. He reminds me of my neutronics teacher who was a physics guy. There's no hate, just an amusement at the unnecessary rigidity when talking specifically about actual nuclear power plant operations.

Just to clarify what I am and what I do, I'm a reactor engineer with experience as an operator. I actually work at a nuclear power plant, participate in the design of cores, participate in and supervise plant startups, and regularly write power transient plans. I'm not exactly talking out of my behind here.

So to us, it's fine and dandy that, in the world of universities, research labs, and the rest of the paper bound world, the single word "critical" is used broadly to incorporate what the functional nuclear power world calls "critical mass" and "critical geometry." Yes, we all understand the theoretical basis for these things because we went to school and learned them and apply said education every day in our work.

Even the companies that write the codes, like Westinghouse, Areva, GE, and some now gone or absorbed, older vendors, don't ever use criticality in the fashion that Morbius is so stuck on. We always specify "critical mass" or "critical geometry" if you want to talk about those specific aspects of core design. In the world of nuclear power, which I have restricted my comments to, criticality is always in terms of neutron population.

To talk about who is using what terminology in a "non-standard" fashion is needless. To a physicist, it's purely in the mathematics. They will use "critical" in the abstract mathematical sense, which considers only the mathematical model. In the functional nuclear power world, "critical" is always in regards to actual neutron population behavior.

This statement
Morbius said:
The above shows that you really DON'T understand some very important features about
the operation of a nuclear reactor. I would defy you to calculate reactor transients,
and do reactor "noise" analysis with your concept of what criticality is.
... is laughable. I understand it just fine. I just happen to be in the position of having to deal with the reality of nuclear power operations, rather than just spouting theory and mathematics at people.

Maybe if I go to the nearest nuclear engineering grad program, and get my PhD, my true-to-life experiences will stop meaning anything? Maybe it has to be a physics PhD...

quetzalcoatl9
It's like the old adage from the aerospace industry; "Planes are designed by people with PhD degrees, built by people with Master degrees..."

Morbius,

I couldn't agree more.

As someone currently doing their PhD dissertation in theoretical chemistry, I know what you mean. I worked in industry for several years before going to grad school, so I am well familar with the false assumptions behind the attitude that Emfuser is espousing. I've even been referred to once as "too academic", as if that were an insult :rofl:

Despite being a "paper-bound scientist" I have managed to design a few systems at my computer-bound desk that have worked out just fine in the non-paper-bound "real world" (as if the entire theoretical physics established by humanity were somehow "not real") My experimental collaborators seem happy enough at these abstract, and hence clearly false, results. :tongue:

quetzalcoatl9
Maybe if I go to the nearest nuclear engineering grad program, and get my PhD, my true-to-life experiences will stop meaning anything? Maybe it has to be a physics PhD...

No, your experiences of course mean a great deal and there is no substitute for that. But you may also realize that folks who have a PhD in physics have worked very hard to understand their subject matter extremely well.

It was the endeavors of academic physicists that resulted in the development of nuclear science (over 90% of all cutting edge science is still done at universities and government labs) - in a setting of peer review, not supply and demand. Do you respect that even a tiny bit?

Emfuser
Morbius,

I couldn't agree more.

As someone currently doing their PhD dissertation in theoretical chemistry, I know what you mean. I worked in industry for several years before going to grad school, so I am well familar with the false assumptions behind the attitude that Emfuser is espousing. I've even been referred to once as "too academic", as if that were an insult :rofl:

Despite being a "paper-bound scientist" I have managed to design a few systems at my computer-bound desk that have worked out just fine in the non-paper-bound "real world" (as if the entire theoretical physics established by humanity were somehow "not real") My experimental collaborators seem happy enough at these abstract, and hence clearly false, results. :tongue:

I believe you misinterpret my words.

I have a real problem with people stuck in the theoretical world who go around beating people over the head with their degrees. I have an even bigger problem with the people who not only beat people over the head with their degrees, but they do so to those who are actually knee deep in the applied world, and not sitting at a desk or in a lab.

It is pretentiousness such as what I see in this thread that causes the very disrespect that you describe. The failure of high-horse-riding scientists to acknowledge the realities of how something is actually done vs. the conceptual world is the very reason that the idea of being "too academic" exists. Any engineer recognizes full well the value of the theory and methodology we are taught, but it's not until you take that engineer, slap a hardhat and some PCs on him, and drag him into containment while we're offline does he or she gain the understanding of what is actually going on when you have to take those ideas you came up with on paper, and try to fit them with reality.

Thus, I will reiterate that I do not come into this thread with any agenda to disrespect, but I have little tolerance for people who don't actually do the job I do, telling other readers that I have a gross misunderstanding of the concepts at hand while simultaneously making needlessly obnoxious comments about academic degrees, and figuratively thumping their chests, with authority.

That turbine sitting a few hundred feet away from me is still spinning. That reactor is critical. No, we don't have any silly requirement about having a neutron source in the core all the time.

Now, can we agree to acknowledge that there are two schools of equally valuable thought here? One is the purely theoretical, one is applied. They each have their own nuances, and it would probably be a good idea for those involved in each to be representing their respective schools of thought, without one trying to put on the hat of the other, and speak for them.

Is such a suggestion agreeable to you academic folks? You can talk in the realm of theory all you want, but instead of trying to stifle and browbeat people who come from the applied side, actually consider their viewpoints.

:)

Dearly Missed
Not really

Morbius is the typical paper-bound scientist.

WRONG!!! I have worked on design and design methods for REAL devices; real
nuclear reactors, and other real nuclear systems.

These reactors and systems have been BUILT and TESTED!!!

NO - you can't paint me as someone who is in an ivory tower somewhere.

My experience is AT LEAST as "real" as yours!!!

Just to clarify what I am and what I do, I'm a reactor engineer with experience as an operator. I actually work at a nuclear power plant, participate in the design of cores, participate in and supervise plant startups, and regularly write power transient plans. I'm not exactly talking out of my behind here.

Could have fooled me!!!

So to us, it's fine and dandy that, in the world of universities, research labs, and the rest of the paper bound world, the single word "critical" is used broadly to incorporate what the functional nuclear power world calls "critical mass" and "critical geometry." Yes, we all understand the theoretical basis for these things because we went to school and learned them and apply said education every day in our work.

Even the companies that write the codes, like Westinghouse, Areva, GE, and some now gone or absorbed, older vendors, don't ever use criticality in the fashion that Morbius is so stuck on.

WRONG!!!! My thesis supervisor was the late Prof. Allan F. Henry - who used to be
the head of codes and methods development at Westinghouse. Prof. Henry was a
stickler for the proper use of these terms. In his textbook, which is one of the seminal
texts in nuclear reactor theory; Prof Henry made a point of the fact that "critical" is a
property of the geometry and materials, and does NOT mean neutrons!!!

We always specify "critical mass" or "critical geometry" if you want to talk about those specific aspects of core design. In the world of nuclear power, which I have restricted my comments to, criticality is always in terms of neutron population.

And you are WRONG!!! It is those "in the industry" that have BASTARDIZED the term;
and use it incorrectly.

I note that you sidestepped my challenge to answer the questions I posed in my last
post. It would have been illustrative for you to have attempted them.

However, let's consider a super-critical assembly - like the example I gave of the
core of Little Boy before it exploded. Let's assume that we have this assembly sitting
there without any neutrons present.

Now suppose I introduce a single neutron heading into the assembly. [ This is the type
of problem nuclear weapons designers have to consider.] The interactions of neutrons
with the material is stochastic - that is it is probabilistic - NOT deterministic. There is
a finite probability that the neutron thus introduced will be absorbed without starting a
chain reaction. That's because the radiative capture cross-section of the Uranium is
finite.

All we can say is that there is a certain probability that the neutron so introduced will
lead to a chain reaction - it is not guaranteed. So how do we determine that probability?

That probability is a VERY REAL physical property of the assembled nuclear system.

It turns out - that the probability is a function of the "k"; or "criticality" of the assembled
system - something Emfuser claims doesn't exist because there are no neutrons.

There's a whole inter-related body of physics and knowledge here that requires that
"k" and "criticality" exist as a function of the geometry and materials independent of
whether there are neutrons present or not.

Emfuser, you are analogous to the child that is playing a toy piano wherein all the black
keys are just painted on. The only notes you have at your disposal are those in the key
of C-Major. The problem is you don't see any use for the other notes.

Well - there is a WHOLE WORLD of music and music theory beyond the key of C-Major.

There's the key of D, E,....as well as the existence of "minor keys", and "diminished keys"
... ALL of which you see no use for because you are only familar with the notes in the
key of C-Major.

Then you would have the audacity to tell Mozart that he doesn't need any notes beyond
those of C-Major; when Mozart has actually composed symphonies, that you HAVEN'T!!

... is laughable. I understand it just fine.

You keep proving that you DON'T understand just fine.

I just happen to be in the position of having to deal with the reality of nuclear power operations, rather than just spouting theory and mathematics at people.

I doubt that you have done say a reactor noise analysis? Have you ACTUALLY done
reactor noise analysis?

OK - you're a startup engineer. Did you DESIGN the reactor?

A friend of mine was a startup engineer. I'm familiar with the job. You've essentially
been given the reactor - you DIDN'T design it. You didn't deal with all the physics that
require the mathematics and concepts that I'm talking about.

Maybe if I go to the nearest nuclear engineering grad program, and get my PhD, my true-to-life experiences will stop meaning anything? Maybe it has to be a physics PhD...

If you do go study at a nuclear engineering grad program; you would find out that
"criticality" is a property of the geometry and materials.

Dr. Gregory Greenman
Physicist

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Dearly Missed
Btw, I would be interested to read your input on the questions you posed on the example scenarios.
curie,

Since Emfuser took a pass at answering the questions i posed; I'll do so for you.

In regards to the core of the Little Boy device sitting there without any neutrons:

Is Little Boy critical? Yes - in fact the assembled Little Boy core is super-critical.
Even though there are no neutrons, the "k" of Little Boy is defined; and it is greater than
unity. The "k" and the "criticality" are properties of the geometry and materials;
irrespective of whether there are neutrons in the device at the time.

As I explained in my previous response to Emfuser, one of the physical properties
of the super-critical assembly is the probability that a neutron introduced at a given
position, with a given energy, and going in a given direction will lead to a chain reaction,
as opposed to immediately being absorbed, or having the chain reaction start and then
die out. It turns out that the probability that the neutron will cause a runaway chain
reaction is a function of the "k" the assembly [ in addition to the properties of the added
neutron].

Under Emfuser's understanding [ or in actuality lack thereof ]; there would be no "k"
defined because his "understanding" requires neutrons to be present. The super-critical
Little Boy assembly has a physical property every bit as REAL as any other physical
property - namely this probability - but we can't characterize it without the value of "k".

Evidently Emfuser is also ignorant of the concept of "neutron importance". You need
the concept of neutron importance when you define reactivity. Neutron importance
will properly account for the fact that a given number of neutron introduced into the
core will have differing effects on the reactivity based on the position, energy, and
direction of the incident neutrons.

The neutron importance is the solution to the adjoint transport equation. However,
unlike the forward transport equation which solves for a physical quantity - namely
the distribution of neutrons in phase space - there are no "adjoint neutrons". So
Emfuser would contend that it isn't "real".

As another analogy, consider the suspension system on your car. This suspension
system can either be underdamped, overdamped, or "critically damped". [ If your
"shocks" are worn - it will be underdamped. ] This property of the car's suspension
system EXISTS and is WELL DEFINED - independent of whether the suspension
system is actually moving at the time. The car's suspension is either underdamped,
overdamped, or critically dampded; even if it is not moving. Likewise, a reactor is
sub-critical, super-critical, or exactly critical; independent of whether there are actually
neutrons in the reactor multiplying or attenuating.

Emfuser knows only basic Reactor Theory 101. Fortunately, I would assume the
people that actually designed the reactor he works on have a more comprehensive
knowledge of Nuclear Reactor Theory.

The question about putting a pulse into an exactly critical reactor:

Putting a pulse of neutrons into the reactor DOESN'T CHANGE the criticality.

Likewise, putting a pulse of neutrons in the reactor DOESN'T CHANGE the "k" of the
reacor. [ Assuming no heating that would alter the material properties. There will be
an insignificant amount of heating - depending on the magnitude of the pulse ].

The reactivity is different. The reactivity DOES DEPEND on the distribution of neutrons
in the reactor.

Before the pulse. the exactly critical reactor had a reactivity of ZERO. The reactivity
vanishes because the geometry and materials are in an exactly critical configuration,
AND I also premised the question with the fact that the neutron population present was
the fundamental mode. Under those conditions, the reactivity vanishes identically.

However, when I add the pulse of neutrons; I'm augmenting the neutron population with
a bunch of neutrons that are NOT in the same distribution as the fundamental mode.

With the addition of this neutron pulse; the reactor will go through a transient on its way
to a new steady-state. During this transition the reactivity is NOT ZERO!!

When the reactor reaches a new steady state with an increased neutron population; that
new population will be a multiple of the fundamental mode; so that the neutron population
is once again in the fundamental mode and the reactivity vanishes exactly.

[This is all modulo the assumption of constant material properties; i.e. insignificant heating
or the assumption that the cooling system removes any excess heat ]

This is EXACTLY the type of questions one would expect in a graduate program in
nuclear engineering; in a reactor theory or transport theory course.

Dr. Gregory Greenman
Physicist

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quetzalcoatl9
i have a copy of Arfken's "mathematical methods for physicist" from a grad course i took.

in one chapter it gives what is called the neutron diffusion equation as:

$$-D \grad^2 \phi(r) + k^2 D \phi(r) = Q \delta(r)$$

where $$\phi(r)$$ is the neutron flux, $$Q\delta(r)$$ is the source, k and D are constants (I assume D is the diffusion constant).

(the solution involves Fourier transforming, solving, then back-transforming).

i am curious, is the "k" in this expression the same "k" that you guys are talking about? (i know virtually nothing about nuclear science).

theCandyman
Then, can you define "k" for anything? The entire earth, for example?

quetzalcoatl9, that form is different from what I remember. I think that k2 is a coefficent of absorption, but don't quote me.

Dearly Missed
i have a copy of Arfken's "mathematical methods for physicist" from a grad course i took.

in one chapter it gives what is called the neutron diffusion equation as:

$$-D \grad^2 \phi(r) + k^2 D \phi(r) = Q \delta(r)$$

where $$\phi(r)$$ is the neutron flux, $$Q\delta(r)$$ is the source, k and D are constants (I assume D is the diffusion constant).

(the solution involves Fourier transforming, solving, then back-transforming).

i am curious, is the "k" in this expression the same "k" that you guys are talking about? (i know virtually nothing about nuclear science).
quetzalcoatl9,

Afraid not.

$$-D \grad^2 \phi(r) + \Sigma_a \phi(r) = {1 \over k} \Sigma_f \phi(r)$$

The above is a zero-scattering version of the diffusion equation. [ The "sigma a" term
means absorption. If there were scattering, this would be replaced by the total
cross-section and there would be an in-scattering term on the right ] The scattering
just complicates things - so for illustration - let's assume zero scatter.

The term with sigma_f on the right is the fission source.

The inverse "k" multiplier is a way to make the equation "artificially critical".

What we have here is a time-independent form of the diffusion equation. If a system
is really sub-critical, or super-critical; it is not time independent. The neutron flux should
either be falling off for a sub-critical system; or soaring up for a super-critical system.

An equation implies equality or balance. If the system is sub-critical, then destruction
exceeds production. If the system is super-critical, then production exceeds destruction.
So there's no equality, and no equation without a time-dependent derivative term.

However, you can make the system balance artificially by dividing the fission source
on the right hand side by "k". If production is too large; bring it down by dividing by
a "k" greater than 1. If destruction exceeds production; artificially increase the fission
source on the right by dividing by a "k" less than 1 to achieve balance.

Thus you get k > 1 for a super-critical system, and k < 1 for a subcritical system. If the
system is exactly critical - then k = 1.

That's the "k" that we are talking about. It's a eigenvalue of the equation; or equivalently
its inverse is the actual eigenvalue. Note that the equation is homogeneous. That is
you can put all the terms involving phi and its derivatives on one side and the right hand
side will then be zero.

If the equation were non-singular there would be exactly one solution. Well since the
equation is linear and homogeneous; there is an easily seen solution - phi = 0.

Well phi = 0; meaning no neutrons is not what you want for an operating power reactor.

So for particular values of the eigenvalue 1/k; you can make the above equation singular.
That means there will be multiple solutions. With multiple solutions, you can have both
the trivial solution phi = 0; and some more interesting solutions. In fact, you get
INFINITELY many solutions, because if phi is a solution, then any constant multiple of
phi is a solution.

That corresponds to the fact that in a reactor, if phi is the solution; then a constant
multiple is also a solution - corresponding to a power level that is that same constant
multiplied by the power for your original solution. It means a critical reactor can be
run up or down to whatever power you want, as long as you can hold the material and
geometric properties constant.

Dr. Gregory Greenman
Physicist

theCandyman
So if I am understanding this correctly, forcing the equation to be excatly critical makes it time independent? That sounds a little too easy, isn't it just ignoring the actual situation?

Dearly Missed
So if I am understanding this correctly, forcing the equation to be excatly critical makes it time independent? That sounds a little too easy, isn't it just ignoring the actual situation?
Candyman,

No - because what you ultimately WANT is a time-independent equation.

What you ultimately want for your reactor is for it to sit there at a steady-state power level
and just turn out energy.

So in effect - you postulate the end solution that you want - one that is "stationary" - that
is time-independent.

As I stated above; you can only get that if you make the equation "singular".

The system will only be singular for some special choices of the cross-sections and
geometry.

That is - if you are designing your reactor - you adjust the design of the system
so that you eventually get a stationary system with k = 1 ; i.e. the system is exactly
critical.

Now in actual fact; because we don't know cross-sections exactly, and some other factors
it turns out that the system may actually go critical at some different, non-unity value of
"k". Therefore, "k" also provides a systematic way to deal with the errors in your nuclear
data.

If the reactor actually goes critical with k = 1.002; because of small errors in your data;
then the next reactor you design; you shoot for k = 1.002 instead of exactly k = 1.

Dr. Gregory Greenman
Physicist

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quetzalcoatl9
i see.

does one go about solving this equation iteratively (much like a self-consistent field equation - i assume the actually boundary conditions can be complicated) as opposed to an exact matrix inversion?

theCandyman
I don't think so. I had to use this equation some in my fundamentals class. The one we used was time independent and for monoenergetic neutrons. It didn't strike me as anything difficult, but I never REALLY understood what it was I was looking at after getting the answer.

As for the boundary conditions, it depends on the geometry. If I remember correctly, sphereical geometry was the easiest with which to deal.

curie
Thanks for the enlightenment on your questions Morbius - gets the old grey matter churning!

I must say I have never had an appreciation of the concept of k, criticality, etc in the purely mathematically derived sense, although talking around the subject here I can see it. I have some brief reactor theory academic training & interact with masters students study reactor engineering & I am confident that none of these students over the last couple of years will have a clue about this level of the theory. But on the other side, the practical side, they can be equally clueless, not having done any lab work since their first undergraduate year, eg no obvious appreciation of the basic concept as to why one would take more than one reading of a varying parameter & compute an average, not being able to interpret a log scale to take a reading from a chart, etc

Onwards from student level, I don't think it is necessarily a problem - broadly speaking there are & always will be operational types & more theoretical types of person as demanded by the world beyond the teaching stage. Ultimately the most theorectical & academic mathematically-based understanding is the highest form and will always be "correct", correct meaning adhering to the most accepted advanced theory. In other different (& thus it could be said lower) levels of understanding, some concepts may be "understood in a different way" & are actually incorrect, but nevertheless they are generally adequate for the model they are applied to, eg having a full understanding of reactor physics theory as opposed to having a more operationally-based grasp. The fullest level of understanding will always work downwards, ie it can be applied to all models & situations, but if you are looking upwards from a different level of understanding, this can result in a problem. I see this often at plant nuclear/reactor safety meetings, where the committee is a mix of both types of people. Ideally it should work collaboratively as a team, getting the best out of all but all too often it ends up in academic vs operational type debates, sometimes this results in hopelessly over-complex & impracticable measures being imposed with little or no operational benefit. On no account though, should disrespect for any viewpoints be allowed to cloud the issues - this is the science & technology field, not politics! So back to increasing our understanding, where it is helpful to discuss not only the "purist" understanding but how it can be understood where it applies in the more operational setting.

Dearly Missed
i see.

does one go about solving this equation iteratively (much like a self-consistent field equation - i assume the actually boundary conditions can be complicated) as opposed to an exact matrix inversion?
quetzalcoatl9,

Yes - the solution is solved iteratively.

It actually requires a "nested iteration" - there's an "inner iteration" and an "outer iteration".

When you solve the equation, you start out not knowing either the flux, phi; nor the
eigenvalue "k" that you are trying to solve for.

Therefore, there is an "outer iteration" that converges on the eigenvalue "k". That
iteration takes the form, typically as a "power iteration"; which you can look up in any
text on iterative methods.

Within each outer iteration, the value of "k" is fixed; but you still have an iterative problem
solving for the value of the neutron flux as a function of space and energy.

As for boundary conditions; "vacuum" boundary conditions are the norm. That is at
the edge of your problem - at the reactor boundary, or beyond the shield if that is included
[ which it should be because it can scatter neutrons back to the reactor ]; you impose
a boundary condition that says there are no incoming neutrons.

Dr. Gregory Greenman
Physicist

Dearly Missed
Ultimately the most theorectical & academic mathematically-based understanding is the highest form and will always be "correct", correct meaning adhering to the most accepted advanced theory. In other different (& thus it could be said lower) levels of understanding, some concepts may be "understood in a different way" & are actually incorrect, but nevertheless they are generally adequate for the model they are applied to, eg having a full understanding of reactor physics theory as opposed to having a more operationally-based grasp. The fullest level of understanding will always work downwards, ie it can be applied to all models & situations, but if you are looking upwards from a different level of understanding, this can result in a problem.
curie,

I agree with your summary above. The fullest and most complete understanding is the
scientific and mathematical understanding. I also see that one can have a "working
understanding" that serves a more limited purpose. That, I believe; is where Emfuser
is.

That's why I analogized his understanding to a musician that only plays in the key of
C-Major. You can make a lot of music in the key of C-Major, and if that's what you
play in; fine, so be it.

However, it is limiting. There's a whole body of music and music theory that goes
beyond the key of C-Major. So Emfuser tellinig me that a device can't be super-critical
unless it has neutrons in it; is like a "musician" telling a composer that there's no note
between "F" and "G". The composer counters with, of course there is; "F#".

Then the "musician" comes back and says that "in the real world" nobody uses "F#".
That's just something to write on the chalkboard. Nobody out making music and selling
records uses "F#".

Obviously, there's a very good use for "F#", and all the other notes that are not in the
key of C-Major.

If this were a forum about Music; one wouldn't want the contention to stand unchallenged
that there was no need for "F#" and the other notes beyond the key of C-Major.

I think one's understanding should be grounded in a complete understanding of the
subject. I hear too many engineers that just "plug and chug"; they plug values into some
formula they learned in school, without an understanding of what the formula really means.

There are many cases where someone used the wrong formula; or forget that the formula
is an approximation and forget under what conditions the approximation is valid.

I recall NASA had some trouble a while back where they lost [ crashed ] a mission
because someone used an approximation that wasn't valid. I would sure hate to have
something like that happen in the nuclear field.

Dr. Gregory Greenman
Physicist

theCandyman
It actually requires a "nested iteration" - there's an "inner iteration" and an "outer iteration".

When you solve the equation, you start out not knowing either the flux, phi; nor the
eigenvalue "k" that you are trying to solve for.

Isn't flux a given measurement?