I was wondering today, how long boron control rods remains in a nuclear power plant? When a boron atom absorbs two neutrons, it becomes the unstable isotope boron-12 and and the boron nuclei starts decaying. After some time it looses two protons and two neutrons (alpha-decay), so it decays into lithium and is not boron anymore.
So, how long does it take until a control rod is completly depleted (no absorbing boron atoms left), that they are useless and have to be exchanged? And I also don't know why they use boron for control rods? Why can't you use any other stable element, because boron is also unstable when it absorbs up to 2 neutrons. What is the special with boron?
[Physics] How long remains a boron control rod
absorptionneutronsnuclear-engineeringnuclear-physicsprotons
Related Solutions
What is the relation between control rods surface exposed into a nuclear reactor and neutron energy?
You mention control rod surface. Why? Do you understand how control rods work? They're inserted into the core and absorb neutrons. Now, why would the surface be a metric that matters? Perhaps this was just the most obvious thing to you at the time, so let me get into the complications here.
Cross sections are higher at thermal versus fast energies. The fast and thermal neutron flux both matter a great deal for typical light water reactors. The main difference is that the average path length of a neutron (before interacting) is much longer for fast neutrons than thermal neutrons. Now, since we've made this distinction, we can ask if the control rods absorb a significant fraction of the neutrons incident on its surface. The answer is mostly "yes" for thermal energies and "no" for fast energies. See, because the thermal flux has a much smaller mean free path, it is much bumpier because of the presence of different materials, including fuel, moderator, and absorbers. It would be mostly correct to suppose that some significant fraction of a thermal neutron beam is absorbed after entering the control rod surface in terms percentage. As an order of magnitude guidance, it would be more than 1%. The same is probably not true for the fast flux. Furthermore, according to transport/diffusion physics, if you were to reduce the fast flux by a significant percentage, you would necessarily suppress the fast flux in the entire region of the core.
Anyway, let me address your question by saying that under the unphysical assumption that a control rod absorbed all neutrons incident on its surface, its reactivity contribution would be proportional to its surface area, relying on a few other assumptions as well. Notably, if you have too many control rods in the same small area they will depress the flux around each other so its not 100% valid. As I've already pointed out, this doesn't fit current reactors.
Let's look at the other extreme. Say, for instance, that the control rods absorbed a negligible fraction of the flux relative to the total flux. This is much closer to normal reactor conditions. In this case, the reactivity contribution would be proportional to the volume of the control rods times the local flux value, and this is shown mathematically in several nuclear engineering text books using perturbation theory.
Is it linear?
Again, use perturbation theory and it's linear for small movements. But what does this physically correlate to? My answer is that it is with respect to the total amount of absorbing material times the local flux where it exists. This is because the control rods don't significantly affect the shape of the flux. In the thermal energies, however, this is a mediocre assumption, which is why people use actual computer codes to design reactors.
(math warning) variables:
- Neutron flux $\phi$
- Absorption cross section $\sigma$
- Absorber atom number density $N$
- Rod cross-sectional area $A$
- Linear distance rod is moved $\Delta l$
Let's speak of an isotropic neutron flux of a single energy. Then the flux will be of a certain physical shape $\phi(\vec{r})$ and the entire point of using the words "perturbation theory" is to refer to the assumption that $\phi(\vec{r})$ doesn't change with the insertion of the control rod. If the end of a control rod is at some given $\vec{r}$ in the core, then the change in reactivity due to movement of the rod will be $\phi(\vec{r}) N \sigma A \Delta l$ which will come out to units of $1/s$, representing the number of neutrons absorbed per second from the volume of the absorber element introduced. This affects the power of the core through a concept called the multiplication factor, which is how much the neutron chain reaction grows or declines for each neutron generation.
$$k = \frac{ \text{Fission neutrons created} }{\text{Neutrons born} }$$
The absorber removes neutrons from the population that could cause a fission and thus create move neutrons, so it can be said to subtract from the numerator of that equation. The reactor changes flux (and thus power) over time as:
$$\phi(t) \propto e^{t \frac{k-1}{\Lambda} }$$
You need not concern yourself with the details other than the fact that inserting absorbers causes the flux to decrease over time.
(end math)
I mean, how do neutron absorbing rate change with the progressive immersion of control rods into the core?
This is the most objective part of the question.
According to my prior arguments, the differential reactivity contribution depends on the flux value at the end of the rod being inserted.
So let me go back to the linearity question. The reactivity contribution from control rods is highly nonlinear with respect to control rod position, because the flux in the core changes dramatically with vertical position. AdamRedwine points out, correctly, that the flux is well-smoothed by core design, but that statement is specific to the xy-plane. This is not true for the z-direction, where it is nearly a cosine shape for a PWR, something else for the complicated thermal-hydraulic and neutronic feedback of the BWR.
Here is an example of a differential control rod worth curve.
It is greater than zero at the top and bottom of the core because some neutrons do leave the core, bounce of Hydrogen, and then enter the core again to cause fission.
For any small control rod movement it's linear. Sure. That follows from the fact that the above graph is continuous. The fact also implies that the integral control rod worth curve is smooth.
Short summary
In fact, in typical reactor, neutron needs to travel quite a lot before it initiates next fission, if during these travels it encounters control rod it is "lost" and chain reaction slows down.
Neutron needs to travel because it needs to lose energy (or in other words slow down), this is because modern reactors are designed in such way that fast neutrons wouldnt be enough to support chain reaction (to know why read the rest!). This is a design decision -- you could have a reactor working on fast neutrons -- it just wouldn't be controlable by control rods!
To answer most crucial OP question:
Q: If neutrons travel from the nucleus of one atom to the nucleus of a nearby atom to split it and perpetuate the chain reaction, doesn't that take place within the fuel itself, on an atomic scale?
A: Not really due to the fact that you need to slow down the neutrons, neutrons travel macroscopic paths, and during these travels might be lost to control rod.
Introduction
In typical modern reactors (experimental ones might be different) you do fission by thermal neutrons (thermal means that these neutrons are in thermal equilibrium with the reactor --- that is have the same speed distribution as it should have in working temperature of the reactor --- neutrons produced by fission have much greater speeds). You can have reactor that works on "fast" (non-thermal) neutrons but these are experimental and much harder to control.
Not every interaction between a neutron and uranium nucleous will result in fission, moreover probability of fission depends on neutron energy. Basically the lower the neutron energy is the more likely fission is. See this chart (from wikipedia):
Typical modern reactors
Neutrons produced in fission have high kinetic energy, so before neutron initiates a fission it must lose most of of the energy, so it's free path is quite long (mean free path is length of path that average neutron travels before initiating next fission). Because of that it is improbable that neutron will initiate fission just after it was produced, because it will still have to much energy. Control rods have plenty of ocasions to catch neutrons.
To slow down the neutrons you'll need them to collide with something (like hydrogen atom, uranium atom and so on). However it turns out that when neutrons collide with heavy atoms they tend not to lose energy, they just change direction (this is just basic mechanics not something nuclear related).
Moderator is a material designed to efficiently slow down the neutrons --- this is a material that has a lot of light atoms (water, graphite, helium). Control rods also displace moderator, so neutrons have lower probability of losing enough energy to initiate fission before they escape the reactor.
You have chain reaction if each fission produces exactly 1.0000 neutron that initiates next fission. To "turn off" the chain reaction you don't need to change this value to 0.0000, not even to 0.9000. Actually if you change the value to 0.99 you'll stop the reaction very fast (I couldn't find precise numbers so this might be off a bit). This is due to the fact that average time between each consecutive fission in chain is low. This means that control rods don't need to drastically alter probability of next fission, just a little bit is enough, even some neutrons wouldn't exit fuel capsule, you need to absorb only tiny fraction to slow down (or kill chain reaction).
What control rods are for
Last thing: control rods are designed to control the reactor and keep it in steady state (that is a proper chain reaction). If you need to shut down the reactor (becaue of some emergency) other means are sometimes employed --- but these specific details of these vary. For example, in some reactors you can introduce "poison" that is a material that absorbs neutrons way faster than uranium, but does not fission, and so starves the chain reaction.
Reactors on fast neutrons
OP asked how does atomic bomb work if there is no moderator there. The answer is you can have both atomic bomb and atomic reactor without moderator. Both can have breeding factor equal or greated than one using fast neutrons. Fission crossection (probability that neutron initiates fission) is lower for fast neutrons but it is enough to both have chain reaction and cascade reaction.
Modern reactors are designed this way that they wouldn't work on fast neutrons --- that is: geometry is designed in a such way that to obtain breeding factor equal to 1.0000 you need thermal neutrons.
It us really easy to create atomic bomb --- you just need to amass a lot of material in the same place and it'll blow.
Obtaining chain reaction on fast neutrons is much harder --- mostly because of the reaction speed. Fast neutrons take a lot of time to lose energy, so you have additional time to slow down the reaction. If you use fast neutrons you don't have this additional time so each step of chain reaction is -- in order of magnitudes! -- faster. In this case control rods are not fast enough and wouldn't stop cascade reaction.
People try to create fast neutron reactors, that use much faster control means. One of the design is accelerator driven systems. In this system breeding factor by design is below 1.0 like 0.995 (once again: these numbers are may be wrong but idea behind them is solid) and additional 0.05 comes from an accelerator (or any other neutron source) that injects neutrons into reactor chamber. It turns out that you can control accelerator intensity fast enough to support chain reaction. Such reactors are "hot" research topic in the field (for example because they can work on "burnt" nuclear fuel).
Best Answer
This publication by IAEA is a good summary of considerations relating to control rod (CR or RCCA) material selection and lifetime management.
I was wondering today, how long boron control rods remains in a nuclear power plant?
So, how long does it take until a control rod is completely depleted (no absorbing boron atoms left), that they are useless and have to be exchanged?
And I also don't know why they use boron for control rods? What is the special with boron?
Why can't you use any other stable element? Other isotopes in control rods are used.