What is meant by "local melting"?

In summary: The problem involves thermodynamics and nuclear physics, but as I said, the local melting confuses me.
  • #1
Lotto
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TL;DR Summary
I have this problem and don't understand the task. What is meant by the "local melting"?
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I suppose it means that the gadolinium melts only at one particular location, but I have no idea how to take it mathematically. Could I just say that local melting is when I destroy a bond between two gadolinium atoms? Could I calculate with this?

I just need to understand what I am supposed to do because I am a little bit confused, otherwise I want to solve it myself.
 
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  • #2
I assume it is half liquid and half solid, as ice in water at 0 degree C.
 
  • #3
What is the context of the question?
Are you building an isotope heat generator?

Maybe the generation of heat, at the centre of a sphere of gadolinium, will heat the centre of the sphere to the melting point of gadolinium, 1585 K.

Energy is released by alpha decay throughout the sphere, but some alpha particles may escape the surface. How far will an alpha particle travel through gadolinium?
Thermal conductivity will limit the heat flux from the centre to the surface.
This has become a thermal critical mass computation.
 
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  • #4
Well, this it the whole task. The problem involves thermodynamics and nuclear physics, but as I said, the local melting confuses me.
 
  • #5
Baluncore said:
What is the context of the question?
Are you building an isotope heat generator?

Maybe the generation of heat, at the centre of a sphere of gadolinium, will heat the centre of the sphere to the melting point of gadolinium, 1585 K.

Energy is released by alpha decay throughout the sphere, but some alpha particles may escape the surface. How far will an alpha particle travel through gadolinium?
Thermal conductivity will limit the heat flux from the centre to the surface.
This has become a thermal critical mass computation.
Well, this it the whole task. The problem involves thermodynamics and nuclear physics, but as I said, the local melting confuses me.
 
  • #6
Lotto said:
The problem involves thermodynamics and nuclear physics, but as I said, the local melting confuses me.
Only the centre of the sphere will reach the melting point. That is local.
 
  • #7
Baluncore said:
Only the centre of the sphere will reach the melting point. That is local.
And how big is the centre? I mean how to determine its mass?
 
  • #8
Lotto said:
And how big is the centre? I mean how to determine its mass?
It is the mass of the entire sphere, that is the answer to the question.

"What is the smallest amount of gadolinium 148 needed to put together to cause local melting from the heat generated by its nuclear decay? Assume that
only a decays take place and the material is at room temperature in the air".
 
  • #9
Baluncore said:
Maybe the generation of heat, at the centre of a sphere of gadolinium, will heat the centre of the sphere to the melting point of gadolinium, 1585 K.
So if I understand it correctly, at the centre of the sphere the nuclear decay will happen and will be heating the centre as well. But why it will happen only at the centre? Why not on the surface for example? And the centre is just a point, I still don't understand how to determine its mass so that I can use it for ##Q=cm \Delta T##.
 
  • #10
Lotto said:
But why it will happen only at the centre?
I suspect that with short-range alpha particles, the heating will happen everywhere within the sphere, but since the centre is furthest from the surface, which remains at room temperature, the centre will have the greatest temperature.

The alpha particle production is determined simply by the half-life and the mass present.
What is the average energy released with each alpha particle?
What is the thermal conductivity of gadolinium?
The thermal energy must escape through the sphere's surface, with an area that rises with the square of the radius. The thermal energy is being produced by a mass that rises with the cube of the radius. There must be a limiting mass, when the surface is at room temperature, but only the centre melts.

Here is a conceptual numerical algorithm, that you could replace with calculus.
Start at the centre, with a spherical radius of 1 mm, and at the melting point of gadolinium.
Work out the energy released by that spherical volume, and the drop in temperature as that flux flows the 1 mm, out through that area of that spherical inner surface. You get a lower surface temperature.
Now add another 1 mm shell, which increases the spherical volume, and so total thermal energy flux, but increases the spherical surface area. Compute the temperature drop across that shell.
Repeat that process until the surface of the outer shell has been reduced to room temperature.
The answer to the OP question, is the mass of a sphere, with that final radius.
 
  • #11
Lotto said:
So if I understand it correctly, at the centre of the sphere the nuclear decay will happen and will be heating the centre as well. But why it will happen only at the centre? Why not on the surface for example? And the centre is just a point, I still don't understand how to determine its mass so that I can use it for Q=cmΔT.
in any part of volume heat is generated in constant div Q =g
Total generated ##Q_1 = \frac{4\pi}{3} R^3 g##
Heat goes our from surface ##Q_2= 4\pi R^2 l ## where l is leak of energy per unit area.
So that the temperature is under control
[tex]Q_1 \leq Q_2[/tex]
[tex]\frac{R}{3} g \leq l_{max}[/tex]
The possible maximum of l would give us an estimate. But even beofre that the core is melting. We need restriction of
[tex]\frac{R}{3} g \leq l_s[/tex]
where ##l_s## depends on surface temperature when the core is at melting point toghether with air temperature. Core undertakes the highest tempeature by symmetry.
 
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  • #12
Lotto said:
So if I understand it correctly, at the centre of the sphere the nuclear decay will happen and will be heating the centre as well. But why it will happen only at the centre? Why not on the surface for example? And the centre is just a point, I still don't understand how to determine its mass so that I can use it for ##Q=cm \Delta T##.

I found this list of solved problems on the internet.
http://web.mit.edu/10.302/www/Fall2001/PROBLEMSETS/ps2_solutions.pdf

Problem 3.95 looks very similar to what you need.

I assume the radius that makes the core hot enough to melt is relevant because if the liquid phase has a significantly different density the sphere could explode or implode depending on the change in volume of the core and the mechanical properties of the material. Such analysis would require a more in-depth study but by staying away from that point you could save the effort altogether.
 
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FAQ: What is meant by "local melting"?

What is meant by "local melting"?

Local melting refers to the phenomenon where a small, specific region of a material undergoes a phase change from solid to liquid, while the surrounding material remains solid. This can occur due to localized heating or other conditions that cause the melting point to be reached in that specific area.

What causes local melting to occur?

Local melting can be caused by various factors, including concentrated heat sources (such as lasers or friction), electrical currents, mechanical stress, or chemical reactions. These factors can raise the temperature of a specific region of the material above its melting point, leading to localized melting.

In which industries is local melting commonly utilized?

Local melting is commonly utilized in industries such as manufacturing, materials science, and electronics. For instance, it is used in welding and soldering processes, additive manufacturing (3D printing), and in the fabrication of microelectronic devices where precise control over material properties is required.

How is local melting different from general melting?

Local melting differs from general melting in that it affects only a small, specific area of a material rather than the entire material. General melting involves heating the entire material uniformly until it reaches its melting point, whereas local melting targets a specific region, often using a controlled and concentrated energy source.

Can local melting be controlled and predicted?

Yes, local melting can be controlled and predicted using advanced techniques and tools. Engineers and scientists use computational models, precise energy delivery systems, and real-time monitoring to control the extent and location of local melting. This allows for high precision in processes like welding, cutting, and additive manufacturing.

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