Temperature of a radioactive material

[Methavix]

Hi all!

I have a certain quantity of a radioactive material (e.g. 1 kg) in the open space. How can I calculate the temperature of this material as a function of time?
If it is useful, we can consider a beta decay material.

Thanks a lot.

 

[mfb]

Model the heat flow out of the object (radiation, convection, ...) and the heat production from the decay, then solve analytically if possible or numerically.

 

[Methavix]

If we consider that the material doesn't receive any heat flows by the space, we have only the heat production from the decay itself. But the problem, for me, it's that I cannot define this heat. Do you know a formula? Is it time-dependent?

 

[mfb]

It is just the power released as electrons, assuming it is thick enough to capture most electrons but not thick enough to capture the neutrinos: decay rate (this will go down exponentially with the lifetime of the radioactive material) multiplied by average energy of the electrons.

 

[Methavix]

As you say I easily can have the total energy produced by electrons (because of the decay), but how can I calculate the equilibrium temperature (as a function of time)? I should consider also the black body radiation, right?

 

[mfb]

but how can I calculate the equilibrium temperature (as a function of time)?

Net heat flow out of the material = net heat production in the material.

Blackbody radiation can be a contribution - the only one if your object is in vacuum. In contact to other things, conduction is relevant, in air (or other fluids) convection might be relevant as well.

 

[Methavix]

Thank you!

 

[Khashishi]

If it isn't black, then it will emit less than black body.

 

[Methavix]

Yes, I know this. Thank you.

 

[Methavix]

Hello people, I have again doubts about the same problem.

The radioactive body releases this energy: N(t)*E
where N(t) is the number of decayed atoms as a function of time, and E is the energy released by each atom.

I consider a black body in vacuum and I write the energy balance this way (I consider only radiation, no contacts with other bodies): sigma*T(t)^4*S*t = N(t)*E
where sigma is the Stefan-Boltzmann constant, T(t) is the temperature of the body as a function of time, S is the total external surface of the body, t is time.
The first part of the equation is the result of a time integration of the Stefan-Boltzmann law (multiplied by the surface).
Is that energy balance right?

From that equation I calculated T(t), but the numerical value is very high... around 100,000 K (few changes in time). So I supposed that anything is wrong.
Thanks!

 

[Chestermiller]

Hello people, I have again doubts about the same problem.

The radioactive body releases this energy: N(t)*E
where N(t) is the number of decayed atoms as a function of time, and E is the energy released by each atom.

I consider a black body in vacuum and I assume that all this energy is converted in heat so I write the energy balance this way: sigma*T(t)^4*S*t = N(t)*E
where sigma is the Stefan-Boltzmann constant, T(t) is the temperature of the body as a function of time, S is the total external surface of the body, t is time.
The first part of the equation is the result of a time integration of the Stefan-Boltzmann law (multiplied by the surface).
Is that energy balance right?

From that equation I calculated T(t), but the numerical value is very high... around 100,000 K. So I supposed that anything is wrong.
Thanks!

Time should not be on the left hand side of the equation. This is the equation for the steady state temperature, or for the quasi-steady temperature (if N is not changing too rapidly).

 

[Methavix]

N is changing rapidly, in general.

I have done the integration because Stefan-Boltzmann law gives a power (energy/time), but to calculate the temperature I need an energy balance. So to convert power in energy I have done this... Is it wrong?

 

[mfb]

If it changes rapidly, then your object will change its temperature notably over the process, and you cannot use the assumption of a (pseudo-)equilibrium. You have to take temperature changes into account and you cannot easily integrate over time.

You have to convert your radioactivity to power, yes. This is quite easy as the rate of decays is proportional to the number of existing atoms.

 

[Chestermiller]

If N is changing rapidly, you need to do a transient conductive heat transfer analysis within the object, including both conduction and accumulation of heat. If the thermal conductivity of the material is high enough, the temperature of the object will be uniform, and you can describe the behavior using time dependent ODEs rather than PDEs.

 

[Methavix]

If I use power instead energy, for radioactive decay I have: [N0*lambda*exp(-lambda*t)]*E
where N0 is the total number of atoms of that body and lambda is the decay constant.

Now can I say? that: sigma*T(t)^4*S = [N0*lambda*exp(-lambda*t)]*E

If this equation is correct I can easily calculate T(t). I tried but the result is again about 100,000 K...

Chestermiller, I have supposed a high conductivity of the body in order to have a uniform temperature. Anyway I don't understand how I can write an equation different from the previous one.

 

[mfb]

Can you show us the numbers you are working with? It is hard to guess which approximation is good and which one is not without those.

The formula looks fine. Maybe your material would really evaporate?

 

[Methavix]

Sure :)

E = 8.874*10^-11 J (or about 553.88 keV)
lambda = 1.287*10^-5 1/s
S = 175 m^2
N0 = 1.2577*10^30

 

[mfb]

10^30 atoms releasing 553 keV each with a lifetime of a day is an initial power of 1 TW, the output of 1000 typical nuclear power plants. That will certainly evaporate your material.

 

[Methavix]

Yes, too much power :) But I didn't know if the error was the equation I used or the problem is too much material.
So you are saying that the problem is too much material, right? I have to reduce the total mass in order not to reach this power.
Is the equation I have used in my previous message correct? (decay power = emitted power)

 

[Chestermiller]

If I use power instead energy, for radioactive decay I have: [N0*lambda*exp(-lambda*t)]*E
where N0 is the total number of atoms of that body and lambda is the decay constant.

Now can I say? that: sigma*T(t)^4*S = [N0*lambda*exp(-lambda*t)]*E

If this equation is correct I can easily calculate T(t). I tried but the result is again about 100,000 K...

Chestermiller, I have supposed a high conductivity of the body in order to have a uniform temperature. Anyway I don't understand how I can write an equation different from the previous one.

$$mC\frac{dT}{dt}=N_0\lambda\exp^{-\lambda t}-\sigma T^4 S$$
where m is the mass of the object and C is its heat capacity. This assumes that there is no external source of radiant energy.

 

[mfb]

Yes, too much power :) But I didn't know if the error was the equation I used or the problem is too much material.
So you are saying that the problem is too much material, right? I have to reduce the total mass in order not to reach this power.
Is the equation I have used in my previous message correct? (decay power = emitted power)

It is correct. The unrealistic temperature is simply a result of the unrealistic amount of material. There is no way to assemble so much radioactive material in one place.

 

[Methavix]

Thanks a lot mfb and Chestermiller!
I will use the equation corrected with the term suggested by Chestermiller and with a smaller value of mass.

Regards

 

[Vanadium 50]

Here is a picture of a piece of Pu-238, which has a substantially longer half-life (90 years vs. one day):

It's red hot because of internal heat.

 

[Methavix]

Thank you. In my calculation I supposed to use Na-24 (half life of 15 hours). I didn't think that radioactive materials was so hot because to the decay itself.

 

[mfb]

You certainly did not have 10³⁰ atoms of it. I would expect that the whole Earth does not have that amount of ²⁴Na.

 

[Methavix]

Ok, I understand the error :)

Now I am supposing to have about 1 ton of Na-24 (2.4*10^28 atoms). I imagine that it is again a huge quantity, but it is just an exercise. The most important thing is that now I am using right equations, thanks to your suggestions.
Thanks