How much shielding would be required to protect living tissue from radiation caused by the fusion of p + b11 and the subsequent fissions of the carbon 12 isomer etc? What energies does this process produce? Does it produce gamma rays? What about x rays? What is the dominant form of energy in this reaction? Is it mostly kinetic? Does a 2 gigawatt reactor require twice as much shielding as a 1 gigawatt reactor?
Thanks in advance!
proton boron 11 fusion / fission shielding
The main reaction produces no gamma rays, no neutrons, nothing but high-speed alpha particles (three per reaction, with a total of 8.7 MeV between them). Stopping the alpha particles electrostatically is how the device generates power. Even if the electrostatic collector system failed, the alphas would still be nowhere near capable of penetrating the vacuum vessel wall.
However, according to Wikipedia, about 0.01% of the time the intermediate carbon-12 will fail to fission, and a 16 MeV gamma will be emitted instead. This requires shielding.
There will also be X-rays from ion-electron collisions (bremsstrahlung), and there will be some neutrons from side reactions. The neutrons are by far the more important problem from a shielding perspective, since anything that will stop fast neutrons will stop X-rays.
It has been estimated that in a working p-11B reactor, the neutron yield will be about one hundred millionth of that expected from a comparable D-T reactor. This means that significantly less shielding is required, but since a given thickness of shielding material results in a fractional reduction, not an absolute reduction, and the output is still several orders of magnitude too high for safe human exposure, you still need a fair bit. Maybe three feet of concrete for a 100 MW reactor.
You could alternately use a 6" - 8"water jacket to thermalize the neutrons and absorb them with a layer of 10B. Unfortunately this results in fission of the boron-10, resulting in ~0.5 MeV gamma rays, which then require concrete or lead shielding outside the boron layer, resulting in a compound shield almost as thick as the simple concrete one. Also remember that you still have to mitigate the 16 MeV side-reaction gammas, which boron-10 doesn't help with.
Due to the fractional-reduction thing, a 2 GW reactor would require only slightly more shielding than a 1 GW reactor, since it only has to reduce the neutron flux by another 0.3 orders of magnitude.
However, according to Wikipedia, about 0.01% of the time the intermediate carbon-12 will fail to fission, and a 16 MeV gamma will be emitted instead. This requires shielding.
There will also be X-rays from ion-electron collisions (bremsstrahlung), and there will be some neutrons from side reactions. The neutrons are by far the more important problem from a shielding perspective, since anything that will stop fast neutrons will stop X-rays.
It has been estimated that in a working p-11B reactor, the neutron yield will be about one hundred millionth of that expected from a comparable D-T reactor. This means that significantly less shielding is required, but since a given thickness of shielding material results in a fractional reduction, not an absolute reduction, and the output is still several orders of magnitude too high for safe human exposure, you still need a fair bit. Maybe three feet of concrete for a 100 MW reactor.
You could alternately use a 6" - 8"water jacket to thermalize the neutrons and absorb them with a layer of 10B. Unfortunately this results in fission of the boron-10, resulting in ~0.5 MeV gamma rays, which then require concrete or lead shielding outside the boron layer, resulting in a compound shield almost as thick as the simple concrete one. Also remember that you still have to mitigate the 16 MeV side-reaction gammas, which boron-10 doesn't help with.
Due to the fractional-reduction thing, a 2 GW reactor would require only slightly more shielding than a 1 GW reactor, since it only has to reduce the neutron flux by another 0.3 orders of magnitude.
93143 is correct. Although I might dispute his order of magnitude numbers re: neutrons. I think it is more on the order of a 1E6 reduction. But the difference is not material in terms of the shielding required as 93143 points out.
An example: Suppose a fission reactor is required to reduce the neutron flux at the surface of the shield from 1E12/cm sq per second to 1/cm sq per second. If the shield required to reduce the neutron flux by 1/10th is 6" a flux of 1E12 will require 6 ft of shield. If the flux is reduced by 1E6 to 1E6/cm sq it will require 3 ft of shield. Very helpful. But it is not commensurate with the 1E6 reduction in flux.
An example: Suppose a fission reactor is required to reduce the neutron flux at the surface of the shield from 1E12/cm sq per second to 1/cm sq per second. If the shield required to reduce the neutron flux by 1/10th is 6" a flux of 1E12 will require 6 ft of shield. If the flux is reduced by 1E6 to 1E6/cm sq it will require 3 ft of shield. Very helpful. But it is not commensurate with the 1E6 reduction in flux.
Engineering is the art of making what you want from what you can get at a profit.
If the first reactor bldg. gets built in America it will be measured in feet and inches. And it will be a big improvement for all our homes.Skipjack wrote:Please can people here use a real measurement system?
Feet and inches... I thought we were doing science here and not home improvement.
The plywood forms for the concrete will be 4ft by 8 ft and the stiffeners used will be 2X4s 2X6s and 2X8s.
Sorry about that.
Engineering is the art of making what you want from what you can get at a profit.
Ah yes, the metric system. It is susposed to be more consistant and universal than other systems, and using base 10 should make scaling more straight forward. Unfortionatly the metric system also sufferes from multiple ways of expressing the same thing. It is easy to confuse terms and conversions between the MKS and CGS systyems. Ergs, Joules, dynes,Newtons, Pascals, torrs, K, C, etc...
Dan Tibbets
Dan Tibbets
To error is human... and I'm very human.
I'm Canadian. I used feet and inches because that was what was used in previous discussions about this subject (most of the numbers come from MSimon), and most people understand them, at least where I come from. And anyway I'm trained for nominal functionality with those units, mostly for converting to MKS before solving a problem (2.54cm/in, 0.3048 m/ft, and 2.2046226lb/kg just off the top of my head).
One of my undergrad fluid mechanics classes had a section near the beginning entitled "Of Slugs And Other Ugly Things". (It's 32.17lbm/slug, just so everyone knows)...
Oddly enough, I don't know my weight in kg (or N, shut up) or my height in cm...
The factor of 1e-8 on neutron yield is from rnebel, for the case of a 100 MWe reactor. If I recall correctly everyone was moderately pleased that it was lower than expected, probably due to the nonthermal plasma having a smaller high-energy tail. Of course, it remains to be seen if his calculations are correct...
...don't talk to me about CGS. Especially in relation to electromagnetics. SI is specifically MKS for exactly this reason...
Torr is a grandfathered unit formerly equal to a mmHg. It's not really metric at all. K and °C are easy; they're just like °R and °F but bigger. The only real trick is to remember that it's K, not °K. I'm with you on ergs and dynes, though...
One of my undergrad fluid mechanics classes had a section near the beginning entitled "Of Slugs And Other Ugly Things". (It's 32.17lbm/slug, just so everyone knows)...
Oddly enough, I don't know my weight in kg (or N, shut up) or my height in cm...
The factor of 1e-8 on neutron yield is from rnebel, for the case of a 100 MWe reactor. If I recall correctly everyone was moderately pleased that it was lower than expected, probably due to the nonthermal plasma having a smaller high-energy tail. Of course, it remains to be seen if his calculations are correct...
...don't talk to me about CGS. Especially in relation to electromagnetics. SI is specifically MKS for exactly this reason...
Torr is a grandfathered unit formerly equal to a mmHg. It's not really metric at all. K and °C are easy; they're just like °R and °F but bigger. The only real trick is to remember that it's K, not °K. I'm with you on ergs and dynes, though...
Evidently you are not familiar with American lumber yards or construction industry. If you go to an American lumber yard and ask for a 1.2192 meter by 2.4384 meter sheet of plywood you are going to come away empty handed unless you can translate.Skipjack wrote:LOL, well as long as you guys get your own measurement system right.
I have found that most americans dont know how many inches go into a foot, or ounces go into a gallon...
I think that generally the consensus in the scientific comunity is though, that the metric system is the way to go.
BTW the trouble with changing is that the whole industry is based on traditional measurements. To get even numbers (like 1.25 by 2.50 meters for instance) the sizes of all the lumber will have to change (that means changing all the production machinery) and repairing legacy construction will be a nightmare. It will probably happen some day (I hope) but we are at least 50 years and maybe 200 years from that.
I'm 186.69 cm tall. And I didn't have to calculate. OTOH I have no idea how many Newtons I weigh. Or kg either.
Engineering is the art of making what you want from what you can get at a profit.
Meters and yards are easy if you do not have to bee tooo precise with it.
A yard is roughly 1 meter (well 1 meter is 1.09 yards). So that is not so bad. I usually simply multiply meters by 3 to get a rough estimate in feet.
And to get pounds you multiply the kilograms by 2 (well 2.2 if you have to be more precise... 2 for the ladies
).
Inches... well as I said, most americans dont know how many inches go into a foot. They also can not tell me how many cubic inches make a gallon. These things are very, very simple in the metric system. 1 dm^3 is a liter. That is 10cmx10cmx10cm.
Also practical is that a litre of water roughly weighs (well actually has the mass of) 1kg. Water freezes at 0 degrees C and boils at 100 and so on...
A yard is roughly 1 meter (well 1 meter is 1.09 yards). So that is not so bad. I usually simply multiply meters by 3 to get a rough estimate in feet.
And to get pounds you multiply the kilograms by 2 (well 2.2 if you have to be more precise... 2 for the ladies
).Inches... well as I said, most americans dont know how many inches go into a foot. They also can not tell me how many cubic inches make a gallon. These things are very, very simple in the metric system. 1 dm^3 is a liter. That is 10cmx10cmx10cm.
Also practical is that a litre of water roughly weighs (well actually has the mass of) 1kg. Water freezes at 0 degrees C and boils at 100 and so on...
Which brings up the thing that bugs me most about this supposedly universal "basic" system. One of the three basic quantities is a derived value! We have meters, fine; and seconds, fine; and kilograms. Kilo?? Shouldn't it be the MGS system? Either that or shouldn't they rename the kilogram so that it is the fundamental value (i don't know, the "hunka" or something) and grams would be a millihunka? Having a kilo ANYTHING as the basic UNIT is weird!93143 wrote: ...don't talk to me about CGS. Especially in relation to electromagnetics. SI is specifically MKS for exactly this reason...
40 inches to the meter is good too.Skipjack wrote:Meters and yards are easy if you do not have to bee tooo precise with it.
A yard is roughly 1 meter (well 1 meter is 1.09 yards). So that is not so bad. I usually simply multiply meters by 3 to get a rough estimate in feet.
And to get pounds you multiply the kilograms by 2 (well 2.2 if you have to be more precise... 2 for the ladies ;)).
Inches... well as I said, most americans dont know how many inches go into a foot. They also can not tell me how many cubic inches make a gallon. These things are very, very simple in the metric system. 1 dm^3 is a liter. That is 10cmx10cmx10cm.
Also practical is that a litre of water roughly weighs (well actually has the mass of) 1kg. Water freezes at 0 degrees C and boils at 100 and so on...
An a liter is a quart roughly.
Last edited by MSimon on Sun Mar 08, 2009 3:43 pm, edited 1 time in total.
Engineering is the art of making what you want from what you can get at a profit.
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Billy Catringer
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Skipjack wrote:Meters and yards are easy if you do not have to bee tooo precise with it.
A yard is roughly 1 meter (well 1 meter is 1.09 yards). So that is not so bad. I usually simply multiply meters by 3 to get a rough estimate in feet.
And to get pounds you multiply the kilograms by 2 (well 2.2 if you have to be more precise... 2 for the ladies
Inches... well as I said, most americans dont know how many inches go into a foot. They also can not tell me how many cubic inches make a gallon. These things are very, very simple in the metric system. 1 dm^3 is a liter. That is 10cmx10cmx10cm.
Also practical is that a litre of water roughly weighs (well actually has the mass of) 1kg. Water freezes at 0 degrees C and boils at 100 and so on...
You would be surprised how many Americans are incapable of reading a ruler or a map. The number is depressingly large.
All you really have to remember on lengths is that a meter is 39 and 3/8" long. From that it is simple math to do the conversions. 1m = 39.375"; 1cm = .39375"; 1mm = .039375". A Kilogram = 2.2 lbs. It is the volume measurements that are difficult to remember and deal with off the top of your head.
I do wish, however, that the Metric Gods would make up their pea-pickin' little minds about units like the calorie.
Oh, and let's say that the Watt = 746 horsepower and be done with it, okay? Better yet, Americans, start rating your engines in Watts OUTPUT and be done with it.
Jeez! At least we don't have to fool with pottles or ells or any of the rest of that madness.