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Thought about this from one of the nuclear meltdown threads...

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- Thread starter Arctic Fox
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Thought about this from one of the nuclear meltdown threads...

- #2

Astronuc

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As a first approximation or in large volume (effectively infinite):

One simply needs the decay per unit volume. That provides a heat source [itex]\dot{Q}[/itex]. With the heat source, then one simply solves the heat conduction equation.

[itex]\dot{Q}[/itex] = E * A, where E is energy per decay, and A is activity.

The activity is simply [itex]A = \lambda N[/itex] where [itex]\lambda[/itex] is the decay constant and is defined by [itex]\lambda = \frac{0.69315}{t_\frac{1}{2}}[/itex]. N is the atomic density of the radionuclide.

0.69315 is approximately ln(2).

This ignores the energy lost from particles, particularly gammas that leave the volume of interest. Alpha particles, on the other hand, travel on the order of 1 mm from parent nucleus, and beta particles travel a few mm or cm depending on energy.

=====================

For small finite sources, gamma rays would leave the source and scatter in whatever matter surrounds the source.

There is a special code, MCNP or Monte Carlo Neutron Photon, which does special calculations for the problem stated in the initial post.

One simply needs the decay per unit volume. That provides a heat source [itex]\dot{Q}[/itex]. With the heat source, then one simply solves the heat conduction equation.

[itex]\dot{Q}[/itex] = E * A, where E is energy per decay, and A is activity.

The activity is simply [itex]A = \lambda N[/itex] where [itex]\lambda[/itex] is the decay constant and is defined by [itex]\lambda = \frac{0.69315}{t_\frac{1}{2}}[/itex]. N is the atomic density of the radionuclide.

0.69315 is approximately ln(2).

This ignores the energy lost from particles, particularly gammas that leave the volume of interest. Alpha particles, on the other hand, travel on the order of 1 mm from parent nucleus, and beta particles travel a few mm or cm depending on energy.

=====================

For small finite sources, gamma rays would leave the source and scatter in whatever matter surrounds the source.

There is a special code, MCNP or Monte Carlo Neutron Photon, which does special calculations for the problem stated in the initial post.

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- #3

Morbius

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Arctic Fox said:

Thought about this from one of the nuclear meltdown threads...

Artic_Fox,

The temperature is not dictated directly by the material.

As Astronuc points out - which material you have will determine a heat

generation rate. You also have to know how the material is being cooled -

and the net "resistance" to heat flow.

The material will get as hot as it needs to so that the outflow of heat will

equal the internal heat generation rate.

Thus if you have the material sitting in air - with relatively poor

heat conduction rate to the air - the material will get hotter than if

you immersed it in water - with relatively better heat transfer

characteristics.

Dr. Gregory Greenman

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- #5

Astronuc

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No, the system will reach equilibrium, if one puts an appropriate radiator in the system (presuming this is an RTG in space). Without a radiator, the RTG would reach a high temperature, which might cause it to melt and disassemble itself.

An RTG system (on a spacecraft) radiates heat to space (there is not conduction or convection in a vacuum), with the heat flux (Q") given by

[tex]Q'' = e \sigma (T_r^4 - T_s^4)[/tex], where

[tex]e [/tex]= emissivity (= 1 for ideal radiator),

[tex]\sigma [/tex]= Stefan's constant,

[tex]T_r [/tex]= temperature of radiator, and

[tex]T_s [/tex]= temperature of space

It is know as "Stefan-Boltzmann Law" - find more at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

The heat rejected is then Q = Q"A where A is the surface area of the radiator.

On a large body like a planet, e.g. Earth or Mars, or on a moon, on can use the ground as a heat sink (conduction). On Earth, the atmosphere or water bodies are the heat sink - combination of conduction and forced convection.

An RTG system (on a spacecraft) radiates heat to space (there is not conduction or convection in a vacuum), with the heat flux (Q") given by

[tex]Q'' = e \sigma (T_r^4 - T_s^4)[/tex], where

[tex]e [/tex]= emissivity (= 1 for ideal radiator),

[tex]\sigma [/tex]= Stefan's constant,

[tex]T_r [/tex]= temperature of radiator, and

[tex]T_s [/tex]= temperature of space

It is know as "Stefan-Boltzmann Law" - find more at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html

The heat rejected is then Q = Q"A where A is the surface area of the radiator.

On a large body like a planet, e.g. Earth or Mars, or on a moon, on can use the ground as a heat sink (conduction). On Earth, the atmosphere or water bodies are the heat sink - combination of conduction and forced convection.

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- #6

Morbius

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Arctic Fox said:

Arctic_Fox,

As Astronuc points out - there is ALWAYS cooling.

In the absence of any type of active or intentional cooling - there is

always "radiative cooling" - the hot object radiates heat away as

radiation.

This will ALWAYS limit the temperature.

Dr. Gregory Greenman

Physicist

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Okay, lemme try this again. :)

How hot would each element get if they were just sitting on a wooden table with a thermometer physically touching the top of the material? :D I'm trying to figure out just how much work may be involved in trying to adapt thermocouples to a lower heat source. Someone here meantion using beter TCs, and I am starting to agree.

And I'm not understanding how to remove heat in space; no air? If I remember right, 1)Radiated, 2)Conduction, 3)Convection.

Just "Radiated" into space?

How hot would each element get if they were just sitting on a wooden table with a thermometer physically touching the top of the material? :D I'm trying to figure out just how much work may be involved in trying to adapt thermocouples to a lower heat source. Someone here meantion using beter TCs, and I am starting to agree.

And I'm not understanding how to remove heat in space; no air? If I remember right, 1)Radiated, 2)Conduction, 3)Convection.

Just "Radiated" into space?

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- #8

Astronuc

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On a table in an atmosphere, one would have conduction to table and air, convection with the air, and perhaps radiative cooling, but the radiative cooling would be a small component compared to the other two.

Space is effectively a vacuum, so radiative cooling is all there is. I already provided an equation for temperature, based on heat flux.

To solve the problem, the heat flux needs to be specified, and on can assume about 3-4K for deep space temperature.

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