# The Volume of the reactor and its criticality

The Volume of the reactor and its criticality!!

What will happen to the "Keff" or to the criticality if the volume of the reactor was compressed to one-half its original volume and why ? if the reactor was operating at critical steady-state before changing the volume.

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Astronuc
Staff Emeritus

You're asking if one increases the atomic density by a factor of two, which would also increase N (atomic density) by a factor of two? That is rather physically impossible for a power reactor. Nevertheless, if one can write an expression for keff in terms of N, then one can also determine the dependence of keff on N.

Also, reducing the volume reduces the neutron leakage.

In reality, one can reduce the volume of a reactor design, but one would not increase the mass or atomic density of the materials.

the Atomic mass is not my question , let us say that the reactor is rapidly compressed to one-half its original volume, (maintaining its
shape and keeping the same neutron population as it is compressed)
Will the non-leakage probability increase because the thick of the reactor will increase so the neutron population will increase with time !! and all that will increase the multiplication of the reactor and leave it in a supercritical state ? is this making sense to you ?

QuantumPion
Gold Member

Are you referring to the 65 tons of fissile material in each of the three reactor cores that has collapsed and slumped from it's original configuration and incorporated lots of foreign material as it deteriorated ?
I think he is just imagining a thought experiment, since he is talking about doubling atomic number densities but keeping identical geometry, which is non-physical.

Anyway, non-leakage probability would go up but resonance escape probability would go down. I'm not sure whether k would go up or down or stay the same since there are competing effects. I'll have to think about it for a while.

I think he is just imagining a thought experiment, since he is talking about doubling atomic number densities but keeping identical geometry, which is non-physical.

Anyway, non-leakage probability would go up but resonance escape probability would go down. I'm not sure whether k would go up or down or stay the same since there are competing effects. I'll have to think about it for a while.
Of course I know that physically impossible , let us assume that happen , well we agreed that non-leakage probability will increase , but I'm not sure why the resonance escape probability will go down because mainly it depends on the neutrons' energies inside the core and that won't change even so the change will be small leaving the biggest effect for the non-leakage probability !! but there is an issue that the geometrical buckling will increase of course won't that decrease the non-leakage probability !!? but still its now several times as thick in mean-free paths! so Neutron pop grows exponentially in time. I'm not totally sure about this.

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Astronuc
Staff Emeritus

the Atomic mass is not my question , let us say that the reactor is rapidly compressed to one-half its original volume, (maintaining its
shape and keeping the same neutron population as it is compressed)
Will the non-leakage probability increase because the thick of the reactor will increase so the neutron population will increase with time !! and all that will increase the multiplication of the reactor and leave it in a supercritical state ? is this making sense to you ?
This is not really making sense. First of all, I didn't refer to the atomic mass, but instead to the mass density (kg/m3) or atomic density (atoms/m3). If one compresses matter, one increases mass/atomic density, which for solids and liquids isn't practical for large volumes. This would also mean increasing fission density.

Now if one is simply wanting to decrease the size of a reactor by reducing the amount of fuel and coolant, while keeping the mass/atomic/fission densities the same, then that's a different problem.

From the rapidly compressing a reactor to one-half of original volume, while maintaining shape and mass, then one is doubling the density of the fuel material, structural material and coolant.

So if one can write keff in terms of the macroscopic cross-sections, Ʃ, which is just Nσ, then one could readily solve the problem. I'm not sure how valid calculations of resonance escape probability would be since that and other factors are calculated at normal densities.

If one collapses the solid/fuel material, which would exclude the moderator, then the system would likely be undermoderated.

Astronuc
Staff Emeritus

Of course I know that physically impossible , let us assume that happen , well we agreed that non-leakage probability will increase , but I'm not sure why the resonance will go down because mainly it depends on the neutrons' energies inside the core and that won't change even so the change will be small leaving the biggest effect for the non-leakage probability !! but there is an issue the geometrical buckling will increase of course won't that decrease the non-leakage probability !!? but still its now several times as thick in mean-free paths! so Neutron pop grows exponentially in time. I'm not totally sure about this.
Self-shielding by the fuel.

If everything else stays the same (including the current density at the surface), a small surface area would mean less leakage of neutrons, and so keff should increase.

QuantumPion
Gold Member

Of course I know that physically impossible , let us assume that happen , well we agreed that non-leakage probability will increase , but I'm not sure why the resonance escape probability will go down because mainly it depends on the neutrons' energies inside the core and that won't change even so the change will be small leaving the biggest effect for the non-leakage probability !! but there is an issue that the geometrical buckling will increase of course won't that decrease the non-leakage probability !!? but still its now several times as thick in mean-free paths! so Neutron pop grows exponentially in time. I'm not totally sure about this.
If the fuel is twice as dense, fission neutrons will have more interactions in the fuel before getting out to the moderator. More interactions in the fast region means more resonance absorption.

The greater surface-volume ratio would increase leakage, however the doubled material density would decrease the mean free path. I'm not sure which effect would dominate or if they would cancel out.

To try and provide you with something more tangible:

If in your thought experiment your "reactor" is simply a sphere of fissile material then for a constant keffective, the product of density and system radius remains a constant.
So, increasing the the density of your system, say, from 10 to 100g/cc will require the radius to be reduced by a factor of 10 for it to remain in the same exactly critical state.

If you halve the volume then you will double the density. In doing so, for a sphere, the radius will not halve of couse, it will be perhaps about 80% of the initial value. Therefore, your system will have a keffective > 1.

Considering your system as a reactor with fuel, coolant, moderator etc then, remember, keffective is equal to the product of the fast fission factor, the resonance escape probability, the number of neutrons produced per neutron absorbed in the fuel, the thermal utilisation, the thermal neutron non-leakage probability and the fast neutron non-leakage probability, i.e. the six factor formula which you can find in any good reactor physics text.
This should help you understand the competing processes which determine whether a system can be made critical or not.
Have fun!