nuclear-engineeringnuclear-physicspower
If lead can absorb or block radiation, would it be possible to pump molten lead into a reactor core which is melting, so that it would eventually cool and contain the radiation?
Is there something that can be dumped into the core that will both stop the reaction (extremely rapidly) AND will not combine with radioactive material and evaporate into the atmosphere, thus causing a radioactive cloud?
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:
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.
A nuclear reactor cannot explode like a nuclear weapon.
For a thermal reactor -like Chernobyl or Three Mile Island- the neutron generation lifetime is too long.
For a fast reactor (and a thermal reactor) there is no mechanism for creating and maintaining a super prompt critical assembly sufficiently long for significant release of energy from fission.
You have to really work hard to assemble the correct material to create a nuclear weapon; you need to create a system that is super prompt critical using fast neutrons and remains so sufficiently long for the chain reaction to produce enough energy before pressure causes dis-assembly into a non-critical configuration. By super prompt critical is meant super critical on the prompt neutrons alone without having to wait for the delayed neutrons to contribute. See the Los Alamos Primer by Serber, available from Amazon: the early notes on the physics of a fission weapon from Los Alamos at the beginning of the Manhattan project.
The explosions at Chernobyl were a steam explosion, and a chemical explosion caused by oxygen reacting with aerosolized graphite. (At Three Mile Island the explosion was from oxygen reacting with hydrogen released from oxidation of the over-heated zircalloy fuel cladding.)
For a nuclear reactor, the safety concern is the removal of decay heat from the radioactive decay of fission products after the fission process is terminated. Decay heat is about 5% of full power at shutdown.
A major problem with the Chernobyl design is the core was over-moderated. Specifically, the graphite was the moderator and the cooling water was not needed as a moderator and in fact acted as a neutron poison. So when the cooling water was lost the reactor power increased: positive feedback. The old N reactor in the US was a graphite moderated, water cooled reactor used to produce Pu for the weapons program, but was specifically designed to not have this problem.
Also, Chernobyl did not have a robust reactor containment system, it had a reactor confinement system.
Best Answer
This would guarantee a meltdown.
They're trying to get heat out of the core because---thought the fission chain reaction has been suppressed---various unstable fission daughters continue to decay. Adding hot lead would add heat to the system and not stop this behavior. Total disaster.
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If the core does slag out, it will probably end in a hot (thermodynamically and radiologically) heap at the bottom of the primary containment vessel. Presumably you leave it there to cool a bit, they pour on a lot of concrete and post "Keep Away" signs.
No need to use anything as expensive lead, as you just want to pile on enough mass to suppressed the primary flux. Dirt would do, but concrete will make it hard for stupid people to go digging in the pile.
Adding some boron to the mix would be helpful because otherwise neutrons get everywhere. Sneaky little bastards that they are.