Heat Conduction with a nuclear source

In summary: The Transport Phenomena article by Stewart, Bird, and Lightfoot posts the following equation for heat conduction with a nuclear heat source:Sn=Sn0 [1+b(r/R)^2] where Sn is volumetric thermal energy, r is radius and R is radius of the fuel pellet.The text state that b is a dimensionless positive constant but post no refrence for this equation or what b is based on. Does anyone know where to find this info or what a general magnitude for b is?I'm trying to calculate fuel melt for a loss of cooling accident in BWR with a fuel pellet swelling to the cladding. Thanks!
  • #1
traijan
15
0
In Bird, Stewart, Lightfoot "Transport Phenomena", they post the following equation for heat conduction with a nuclear heat source:

Sn=Sn0 [1+b(r/R)^2]

where Sn is volumetric thermal energy, r is radius and R is radius of the fuel pellet.

The text state that b is a dimensionless positive constant but post no refrence for this equation or what b is based on. Does anyone know where to find this info or what a general magnitude for b is?

I'm trying to calculate fuel melt for a loss of cooling accident in BWR with a fuel pellet swelling to the cladding. Thanks!
 
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  • #2
I should be able to determined the range of 'b'.

It is a function of enrichment and burnup, really a function of the concentration of fissile isotopes, as well as Xe-135, and Sm, Pm isotopes, which are themselves functions of burnup and power level.

This arises because the thermal flux comes from outside the fuel pin, from the moderator. The concentration of fissile isotopes and certain fission products provides a 'self-shielding' for a fuel element. The flux is attenuated from the outside in toward the center of the pin.

At moderate to high burnup, the power density on the periphery of a pellet can be something like 3 times the average or center power density. A simple parabolic fit is a rough approximation. A quadratic or cubic would be better.
 
  • #3
Astronuc said:
I should be able to determined the range of 'b'.

It is a function of enrichment and burnup, really a function of the concentration of fissile isotopes, as well as Xe-135, and Sm, Pm isotopes, which are themselves functions of burnup and power level.

This arises because the thermal flux comes from outside the fuel pin, from the moderator. The concentration of fissile isotopes and certain fission products provides a 'self-shielding' for a fuel element. The flux is attenuated from the outside in toward the center of the pin.

At moderate to high burnup, the power density on the periphery of a pellet can be something like 3 times the average or center power density. A simple parabolic fit is a rough approximation. A quadratic or cubic would be better.

Thats great! I would have never gone down that route. I am trying to do very very rough estimates on Fukushima 3 and what temperture their fuel got to when core cooling is lost. Since little data is still available I am making very broad assumptions.

Core was operatin 180 days, middle of cycle, and shtdown was reached without issue. at the time of cooling loss decay heat in each fuel rod was 7,180 Btu/degF-hr and steam temperture in the vessel was intially 534deg f with no heat removal other then sperheating of steam.

Thus far my b is on order of magnitude of 100 to 1000 but I'm not sure if this is expected or not. (assuming core melt was reached at 2800 deg F)

Do you know if the aboveTransport equation is used in other reports or text? I found the derivation very well written but the lack of refrences limits mein what I can do. I checked out several EPRI manuals n fuel design and the standard Lamrash and Duderstadt refrencesand couldn't find anything for accident transient with no forced cooling. thanks for the help!
 
  • #4
Ah - decay heat would follow the radial burnup distribution in the pellet.

The factor b should be less than 2. I'll try to find a number for a local burnup of ~30 GWd/tU. I expect there are four batches of fuel. Assume typical 8x8 fuel rod geometry.

I doubt the fuel (UO2) temperature got to melting.

The critical parameter is the heat transfer coefficient at the cladding surface. That was probably pretty low due to effectively stagnant, but saturated steam at the liquid/vapor interface, but possibly superheated above. One has to think of pool boiling.

One needs the decay heat curve, and bearing in mind it was cooled initially for about 1 hour (
EDG operating), and then on batteries (for a few hours), then lost the cooling. After that, it went into boiling at some pressure less than operating, but greater than 1 atm.
 
  • #5
I don't think the entire core melted, but don't you feel that some pellets could have reached there melting point in the center? Early reports are showing that Unit 3 may have gone alomst four days without any cooling and the suppression pool torus may have depressurized from either the rupture disks blowing or vessel damage from the hydrogen damage.

Either way, if your estimate for b is correct, my calcs are way off and I need to review my data.

Thanks for your help! Great website!
 
  • #6
Try these data. Pellet Radial Position is normalized. The burnup is in GWd/tU.
The Rel PowDens is not really relevant for decay heat - just the burnup. One can ratio the profile to lower (average) burnups, with reasonable approximation.
Code:
Rad.Pos     Rel     Local
Normal    PowDens  Burnup
0.000     0.8737    31.05 center
0.051     0.8742    31.07
0.114     0.8753    31.12
0.177     0.8771    31.20
0.241     0.8795    31.32
0.304     0.8827    31.48
0.367     0.8867    31.68
0.430     0.8916    31.92
0.494     0.8978    32.20
0.557     0.9054    32.54
0.620     0.9152    32.95
0.658     0.9224    33.23
0.696     0.9311    33.55
0.722     0.9380    33.79
0.747     0.9460    34.06
0.772     0.9554    34.36
0.798     0.9668    34.70
0.823     0.9808    35.10
0.848     0.9984    35.59
0.873     1.0220    36.18
0.899     1.0530    36.96
0.924     1.1000    38.05
0.937     1.1330    38.79
0.949     1.1760    39.75
0.962     1.2370    41.06
0.975     1.3290    43.02
0.987     1.4970    46.54
1.000     2.4190    65.90 OD
          
        Avg Burnup  35.02

Or one can fit with the parabola and get an approximate value for 'b'.
 
  • #7
Learning has occured! Thanks so much for the help and sending me down the right path.
 

FAQ: Heat Conduction with a nuclear source

What is heat conduction with a nuclear source?

Heat conduction with a nuclear source refers to the transfer of heat energy through a material that is heated by a nuclear reaction. This type of heat transfer is commonly used in nuclear power plants to generate electricity.

How does heat conduction with a nuclear source work?

Heat conduction with a nuclear source works by using the heat energy produced by a nuclear reaction to heat a material, such as water, which then turns into steam. The steam is then used to turn turbines and generate electricity.

Is heat conduction with a nuclear source safe?

When performed properly, heat conduction with a nuclear source is considered safe. Nuclear power plants have numerous safety protocols in place to prevent accidents and protect workers and the surrounding environment. However, accidents can still occur, which is why strict regulations and safety measures are in place.

What are the benefits of using heat conduction with a nuclear source?

One of the main benefits of using heat conduction with a nuclear source is that it produces a large amount of energy compared to other energy sources. It is also cleaner than fossil fuels, as it does not produce greenhouse gases. Additionally, nuclear power plants can operate for longer periods of time without needing to refuel, making them more efficient.

Are there any drawbacks to using heat conduction with a nuclear source?

One of the main drawbacks of using heat conduction with a nuclear source is the potential for accidents and the release of radioactive materials into the environment. There is also the issue of nuclear waste disposal, as it can remain radioactive for thousands of years. Additionally, the construction and maintenance of nuclear power plants can be costly.

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