FAQ - D-D fusion - Energy - Flux - Dose
Posted: Tue May 20, 2008 1:55 pm
This effort will discuss only amateur deuterium-deuterium fusion. It will discuss energy releases, how to figure flux, discuss does rate versus equivalent dose.
All calculations will be carried out to only the nearest tenth part of any value.
Deuterium-Deuterium fusion data
Two reactions occur with about equal probability.
D + D = T (tritium 1.01 MeV) + P (3.02MeV)
D + D = 3He (.82 MeV) + n (2.45 MeV)
NOTE* There is a third reaction possible, but it only occurs once in every ~50,000 D-D fusions and that is D+D = 4He + gamma ray.....We will not even consider this reaction in any of our work due to its extreme rarity.
Thus D-D fusion is a mixed output of the above two major reactions such that the energy from D-D fusion (7.3 MeV) shakes out to the following fractions of the net ongoing reactions.
T – 13.8%
P - 41.4%
He3 – 11.2%
n – 33.6%
Figuring energy per particle or number of particles per unit energy
The neutron has an energy of 2.45 MeV
1 MeV of kinetic energy is equal to 1.6x10e-6 ergs
A D-D fusion neutron has 2.45 x 1.6 x 10e-6 = 3.9 x 10-6 ergs of energy
There are 10e7 ergs in a joule, thus, there are 10e7/3.9 x 10e-6 = 2.6 x 10e12 D-D neutrons required to represent 1 joule or 1 watt second of energy.
To obtain 1 watt/second of neutrons we must produce 3 watts of D-D fusion.
Countrawise,
1 watt of D-D fusion only contains 0.336 watts of neutrons or .336 x 2.6x10e12 = 8.7 x10e11 neutrons.
Figuring flux
All calculations are based on an assumed point source of fusion (never the case), in order to simplify calulations.
Only one bit of geometry is needed. The surface area of a sphere is A = 4 pi r^2
Flux is generally considered to be the number of particles, (in this case, neutrons), passing through an area of one square centimeter in one second.
Figure the neutron flux from a 1 watt output D-D fusion reaction at a range of 1 meter.
A sphere of 1 meter radius has an area of 4 x pi x 100^2 = 125,664 sq cm
There are 8.7 x10e11 neutron emitted from this reaction
The flux from this reaction at 1 meter would be 8.7 x 10e11 / 125,664 = 7 x10e6 n/sq cm/second.
The difference in dose rate versus total equivalent dose.
Human tissue differs in its absorption of neutrons based on their energy. A plot exists whereby over a span of different neutron energies one might find what flux will produce a given equivalent dose. All dose rates discussed here are once again for the 2.45 MeV D-D fusion neutron and will differ for neutrons of other energies.
Dose rate
Radiation dosage is quantified by the ‘rem’ or Roentgen Equivalent Man. Neutrons of varying energy have varying effects and a correction chart for tissue absorption of neutrons of varying energy shows that for D-D neutrons about 8.4 neutrons per square cm per second is equivalent to a dose rate of 1millirem/hour.
If a neutron rate meter shows a fusor is putting out a dose rate of 10mrem/hr. Then 84 neutrons are passing through each square cm, each second at the location of the detector probe.
Equivalent Dose
The total dose received over a given period of exposure is the dose rate multiplied by time on exposure. If we are at the probe location in the above example for a period of 8 minutes, we will have received a dose of (8/60) X 10 = 1.3 mrem of fast neutrons.
Another method of arriving at absorbed dose per second can be had when we know the flux at a given point. We suspect that we are receiving a high dose of radiation. We know the flux at our location is 10e5 neutrons per sq cm per sec. How much radiation are we receiving each second? (8.4 x 3600) = 30240 neutrons hitting a square cm to receive 1 mrem/second 10e5 / 3 x 10e4 = 3.3 mrem each second. While not deadly even over minutes of time, a wise man would move out to 10 meters and only receive 3.3/10^2 = 33 micro-rem/sec. A wiser man, still, would construct a simple paraffin-boron shield so that he could stay close to his work.
If we put in 400 watts to our D-D fusor and, by calculation, find we are producing 10e5 neutrons per second isotropic emission, what is the energy input to actual fusion energy output ratio? (This a realistic case of a successful but not optimally operated fusor - typical example)
We know, from above, that 1 watt of D-D fusion emits 8.7 x 10e11 neutrons isotropically each second. So, we are producing 10e5 / 8.7 x 10e11 = 0.1 microwatt of output energy due to neutrons which we have measured. However, This is only 33% of the total fusion energy. the actual total fusion energy emitted is 3.3 X .1 = .33 microwatt. From this total energy figure we arrive at the requested ratio of 400 / 3.3 x 10 e-7 or 1.2 x10e9
Restated we are putting in more than 1.2 billion times as much energy as we are getting out as fusion energy…Or, for each fusion watt output, we must supply to our fusor 1.2 billion watts of electricity.
We find that even the best fusors are still off by a factor of 200 million to one and the average fusor is in the billion to one range. (Nine orders of magnitude improvement to break-even and 11 orders of magnitude before fusion looks attractive as a power source.
So, we can see that by just using the neutron production as measured from a fusor, we can figure the entire fusion energy produced.
It can't be overemphasized that all the simple calcs above have made specific assumptions designed to make the understanding and order of magnitude calcs easy for the newbie or mathematically handicapped. Any step off the assumptions and the results are skewed and can quickly drift off into fantasy if tight results and data are demanded. Fortunately, in the amateur world, we are not called to task on minutia by overlords looking to hold our feet to the fire. Most are here for the effort and not precision. Still, good scientific investigation, amateur or professional, needs to understand the limits of its data collection accuracy and why those limits exist.
Remember that to work with the simplest of calculations, the number of parameters of an experiment must be limited in number and rigidly fixed and held fixed throughout the experiment. If more parameters are introduced which can change, the complexity of the calculations must increase tremendously to deal these new aspects of the procedure or experiment.
P.S. Thanks to many who helped me make such corrections as needed to all of the above with their comments listed in the posts below. All such corrections are now incorporated above as of 5/22/08.
Richard Hull
All calculations will be carried out to only the nearest tenth part of any value.
Deuterium-Deuterium fusion data
Two reactions occur with about equal probability.
D + D = T (tritium 1.01 MeV) + P (3.02MeV)
D + D = 3He (.82 MeV) + n (2.45 MeV)
NOTE* There is a third reaction possible, but it only occurs once in every ~50,000 D-D fusions and that is D+D = 4He + gamma ray.....We will not even consider this reaction in any of our work due to its extreme rarity.
Thus D-D fusion is a mixed output of the above two major reactions such that the energy from D-D fusion (7.3 MeV) shakes out to the following fractions of the net ongoing reactions.
T – 13.8%
P - 41.4%
He3 – 11.2%
n – 33.6%
Figuring energy per particle or number of particles per unit energy
The neutron has an energy of 2.45 MeV
1 MeV of kinetic energy is equal to 1.6x10e-6 ergs
A D-D fusion neutron has 2.45 x 1.6 x 10e-6 = 3.9 x 10-6 ergs of energy
There are 10e7 ergs in a joule, thus, there are 10e7/3.9 x 10e-6 = 2.6 x 10e12 D-D neutrons required to represent 1 joule or 1 watt second of energy.
To obtain 1 watt/second of neutrons we must produce 3 watts of D-D fusion.
Countrawise,
1 watt of D-D fusion only contains 0.336 watts of neutrons or .336 x 2.6x10e12 = 8.7 x10e11 neutrons.
Figuring flux
All calculations are based on an assumed point source of fusion (never the case), in order to simplify calulations.
Only one bit of geometry is needed. The surface area of a sphere is A = 4 pi r^2
Flux is generally considered to be the number of particles, (in this case, neutrons), passing through an area of one square centimeter in one second.
Figure the neutron flux from a 1 watt output D-D fusion reaction at a range of 1 meter.
A sphere of 1 meter radius has an area of 4 x pi x 100^2 = 125,664 sq cm
There are 8.7 x10e11 neutron emitted from this reaction
The flux from this reaction at 1 meter would be 8.7 x 10e11 / 125,664 = 7 x10e6 n/sq cm/second.
The difference in dose rate versus total equivalent dose.
Human tissue differs in its absorption of neutrons based on their energy. A plot exists whereby over a span of different neutron energies one might find what flux will produce a given equivalent dose. All dose rates discussed here are once again for the 2.45 MeV D-D fusion neutron and will differ for neutrons of other energies.
Dose rate
Radiation dosage is quantified by the ‘rem’ or Roentgen Equivalent Man. Neutrons of varying energy have varying effects and a correction chart for tissue absorption of neutrons of varying energy shows that for D-D neutrons about 8.4 neutrons per square cm per second is equivalent to a dose rate of 1millirem/hour.
If a neutron rate meter shows a fusor is putting out a dose rate of 10mrem/hr. Then 84 neutrons are passing through each square cm, each second at the location of the detector probe.
Equivalent Dose
The total dose received over a given period of exposure is the dose rate multiplied by time on exposure. If we are at the probe location in the above example for a period of 8 minutes, we will have received a dose of (8/60) X 10 = 1.3 mrem of fast neutrons.
Another method of arriving at absorbed dose per second can be had when we know the flux at a given point. We suspect that we are receiving a high dose of radiation. We know the flux at our location is 10e5 neutrons per sq cm per sec. How much radiation are we receiving each second? (8.4 x 3600) = 30240 neutrons hitting a square cm to receive 1 mrem/second 10e5 / 3 x 10e4 = 3.3 mrem each second. While not deadly even over minutes of time, a wise man would move out to 10 meters and only receive 3.3/10^2 = 33 micro-rem/sec. A wiser man, still, would construct a simple paraffin-boron shield so that he could stay close to his work.
If we put in 400 watts to our D-D fusor and, by calculation, find we are producing 10e5 neutrons per second isotropic emission, what is the energy input to actual fusion energy output ratio? (This a realistic case of a successful but not optimally operated fusor - typical example)
We know, from above, that 1 watt of D-D fusion emits 8.7 x 10e11 neutrons isotropically each second. So, we are producing 10e5 / 8.7 x 10e11 = 0.1 microwatt of output energy due to neutrons which we have measured. However, This is only 33% of the total fusion energy. the actual total fusion energy emitted is 3.3 X .1 = .33 microwatt. From this total energy figure we arrive at the requested ratio of 400 / 3.3 x 10 e-7 or 1.2 x10e9
Restated we are putting in more than 1.2 billion times as much energy as we are getting out as fusion energy…Or, for each fusion watt output, we must supply to our fusor 1.2 billion watts of electricity.
We find that even the best fusors are still off by a factor of 200 million to one and the average fusor is in the billion to one range. (Nine orders of magnitude improvement to break-even and 11 orders of magnitude before fusion looks attractive as a power source.
So, we can see that by just using the neutron production as measured from a fusor, we can figure the entire fusion energy produced.
It can't be overemphasized that all the simple calcs above have made specific assumptions designed to make the understanding and order of magnitude calcs easy for the newbie or mathematically handicapped. Any step off the assumptions and the results are skewed and can quickly drift off into fantasy if tight results and data are demanded. Fortunately, in the amateur world, we are not called to task on minutia by overlords looking to hold our feet to the fire. Most are here for the effort and not precision. Still, good scientific investigation, amateur or professional, needs to understand the limits of its data collection accuracy and why those limits exist.
Remember that to work with the simplest of calculations, the number of parameters of an experiment must be limited in number and rigidly fixed and held fixed throughout the experiment. If more parameters are introduced which can change, the complexity of the calculations must increase tremendously to deal these new aspects of the procedure or experiment.
P.S. Thanks to many who helped me make such corrections as needed to all of the above with their comments listed in the posts below. All such corrections are now incorporated above as of 5/22/08.
Richard Hull