I Can somebody explain this: Planck's Law in action

[Matthias_Rost]

TL;DR

When you calculate a plasma with 1.000.000 K then you get according to planck's formula a plasma, which emits x-Ray / Gamma-Radiation in a quite deadly intensity. Can somebody explain all this ultra-high temperatures posted in all the fusion experiments?

Plotted is the Irradiance over Wavelength.
Please check for logarithmic scaling.
As you can see, there are 4 curves.
Blue AM 0 as measured
yellow Planck for 5777 K
green Planck for, 5777 K after free space expansion
red Planck for 1.000.000 K

To me the idea of a gamma-Ray-source on earth, below the magnetic field, which protects life on earth from solar radiation, in an intensity, which is way way way outer hand, makes no sense to me. If they really get these high temperatures realized in fusion reactors, we would, according to Planck, already been dead.

 

[Andy Resnick]

TL;DR: When you calculate a plasma with 1.000.000 K then you get according to planck's formula a plasma, which emits x-Ray / Gamma-Radiation in a quite deadly intensity. Can somebody explain all this ultra-high temperatures posted in all the fusion experiments?

Interesting question- one I had not thought of.

First, confined plasmas have a variety of radiation emission mechanisms, not just blackbody but also synchrotron and "bremsstrahlung". There are also emitted neutrons in addition to photons. I did find a brief summary of the safety measures in "Fundamentals of Magnetic Thermonuclear Reactor Design":

"The MFR technological and biological radiation protection is performed by a combination of the blanket and special structural components.
In addition to fusion energy utilisation and tritium breeding, the blanket provides a considerable attenuation of the plasma radiation flux. Physical protection components that absorb neutrons and gamma radiation from nuclear reactions in structural materials are located behind the blanket. Behind the radiation shielding, the parameters of the irradiation effect (particularly onto the magnet materials) must not exceed the following specified limits:
• total absorbed dose for insulators: not higher than 5 × 10⁷Gy,
• nuclear reaction thermal power absorbed by a superconductor: within 10 kW,
• fast neutron fluence on superconducting coil: within 10¹⁹n/cm2.
The shielding material should contain light elements acting as neutron moderators and elements with large atomic numbers absorbing the gamma radiation. The well-reputed heterogeneous iron-and-water medium is generally used for this purpose. Where a thin shielding is necessary, an advanced material based on, for example, zirconium hydride can be used.

In the DEMO and FPP projects, the blanket + shielding thickness is close to 1 m. It is the key component in the gap between the plasma and the TF coil. Approximately 2-m-thick concrete bioshield is used to protect personnel."

Interesting stuff!

 

[Matthias_Rost]

Yeah sure, quite interesting. I met some guys online, who worked at I guess Korean fusion Laboratory. And they said: "We suppress thermal radiation." And I said: "how?". No answer.

See, the main problem in understanding this is sure there are many types of radiation of an ultra hot plasma, when it's rotating in an accelerator. Formally, synchrotron radiation is one kind of decellaration-radiation or called "bremsstrahlung". And yeah, 1 m lead can absorb many of the radiation. But that's not the point. I think there's a misconception. See the Energy, with which they pump the reactor, is one thing. But the true temperature they reach, so the thermal energy is something different. Since thermal energy is defined by Maxwell-Boltzmann-Distribution. And therefore it accounts only relative movement of particles and not aligned movement of particles together in a circle.

So when you now equate the magnetic pumping energy with the kinetic energy derived from Maxwell-Boltzmann, you get quite high temperatures. E_magn = E_kin = 3/2 k_b*T. Truth is, that the relative movement in a plasma of particles, which have a velocity of nearly 1/3 of c in a magnetic field, which is nearly homogenous is almost zero.

That's why I don't believe all these published high Temperature values, unless somebody posts a serious temperature measurement. Doesn't have to be a protocol. But as far as I know after 5000 K every material is at least molten. So the only way to measure the surface temperature of an ultra hot plasma would be optical, like you do at the sun. And how such a spectrum should look according to Planck, I plotted as the red curve.

And I still don't get how to be able to suppress thermal radiation. I mean, this would contradict the 2nd law of thermodynamics. Since, Planck is a quite complex derivative of this fundamental law. So I don't get these guys at all.

 

[Gleb1964]

..
Plotted is the Irradiance over Wavelength.
Please check for logarithmic scaling.
..

The absolute values of irradiance on your graph is applicable only to a enough large volume of plasma which absorb as a Black Body. Small volume would be transparent and have much much less emissivity.

 

[Matthias_Rost]

The absolute values of irradiance on your graph is applicable only to a enough large volume of plasma which absorb as a Black Body. Small volume would be transparent and have much much less emissivity.

The Volume V doesn't matter at all. If you use a tungsten oxide wire and heat it up to 5777 K it will emit almost the same spectrum as the sun. Difference will only be the absorption bands of WO_3 and H/He. And the different prefactors for ε and A. And it's also not about transparency. Transparency would matter if you used a lamp to shine through a plasma. But this plasma is the lamp. Therefore, everything that matters is the surface between the plasma and the vacuum. I = ε*A*σ*T^4. also accounts for gray bodies, when ε<1.

You apparently didn't get Planck's law. For instance, what is ε?. ε(λ) =1-R(λ). So yeah, if the Reflectivity at the interface is greater than zero for the concerning Wavelength, then it's a gray body. Which for perpendicular emission can be written by Fresnel: R =[(n_1-n_2)/(n_1+n_2)]². Vacuum has an n of 1 and most gases have n around 1. And a plasma can sometimes have a higher n, but these plasmas in the reactors mostly have a low pressure, that's why they are not optically dense and that's why n is also small. Example: water --> air. R =[(1,5-1)/(1,5+1)]² =0,04. So (1-R) would be ε = 0,96.

Let's have a look at the blue and green curve. As you can see, there is almost no deviation between Planck using ε = 1,00 and the AM0 Spectrum. Plus, the sun is an optical dense plasma due to its high gravitation. So I would say ε ~ 1,00. The differences between blue and green curve don't originate from Emissivity/Reflectivity issues. These are the absorption bands of He, H, D and T.

See it is the standard procedure to measure temperatures of plasmas optically, because it's the only way. Have you ever seen a measurement of T from a fusion reactor, or do they only publish their calculated values?

 

[Gleb1964]

..You apparently didn't get Planck's law..

What do you think the emissivity coefficient would be if the mean free path of photons (e.g., gamma photons) is much greater than the size of the volume? The emissivity would be about zero. For a black body, the mean free path of photons needs to be much smaller than the volume size, as in the case of a tungsten wire.

 

[Andy Resnick]

The Volume V doesn't matter at all. If you use a tungsten oxide wire and heat it up to 5777 K it will emit almost the same spectrum as the sun.

Yes, but the amount of energy radiated by the wire will be much lower than the sun. The amount of energy you need to absorb to protect you from the amount of UV radiation emitted from a tungsten oxide wire is considerably less than required to block UV from the sun.

 

[Gleb1964]

It’s not the volume itself that matters, but the relationship between the volume size and the photon’s mean free path. That’s what I’m referring to. If radiation passes through a small volume with low absorption, the emissivity of that volume will be proportionally low.