Timeline for How was Enrico Fermi able to predict a power output from his uranium and graphite Chicago Pile 1 (CP-1) reactor?
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| May 4, 2022 at 16:54 | answer | added | Jon Custer | timeline score: 3 | |
| Mar 15, 2022 at 11:23 | history | reopened |
Anders Sandberg Níckolas Alves jng224 Emilio Pisanty Toffomat |
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| S Mar 10, 2022 at 17:04 | review | Reopen votes | |||
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| S Mar 10, 2022 at 17:04 | history | edited | Vogon Poet | CC BY-SA 4.0 |
Focused on specific historical measurement methods
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| Mar 10, 2022 at 13:55 | history | left closed in review |
Jon Custer jng224 BioPhysicist |
Needs more focus | |
| Mar 2, 2022 at 15:53 | comment | added | rmhleo | @VogonPoet there is indeed a way of calculating the energy produced, and he certainly must have used it to calculate the dimensions and amount of material needed. This is a partial differential equation describing sources and sinks of neutrons, called the neutron diffusion equation This a complex equation, but some simple solutions are discussed here. You will see that criticality is defined from the neutron parameters (yield, cross sections, etc.) and that no estimation of heat is needed to estimate performance. Fermi just used neutron detectors. | |
| Feb 28, 2022 at 16:39 | history | edited | Vogon Poet | CC BY-SA 4.0 |
specifically: "Raw"
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| Feb 28, 2022 at 16:11 | history | edited | Qmechanic♦ |
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| S Feb 28, 2022 at 15:10 | review | Reopen votes | |||
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| S Feb 28, 2022 at 15:10 | history | edited | Vogon Poet | CC BY-SA 4.0 |
Modified context to be more generally applicable
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| Feb 28, 2022 at 13:45 | history | left closed in review |
John Rennie jng224 Jon Custer |
Original close reason(s) were not resolved | |
| S Feb 27, 2022 at 22:15 | review | Reopen votes | |||
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| S Feb 27, 2022 at 22:15 | history | edited | Vogon Poet | CC BY-SA 4.0 |
Removed homework-like language and focussed on broad concept of radioactivity with neutron moderation.
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| Feb 27, 2022 at 12:15 | comment | added | NuclearFission | There are two sources of heat. Plain uranium is an alpha emitter, so it will generate heat all by itself from the alpha decay. This will be a very low number since the half-life of uranium is so long. The uranium was purified, so you only have to consider the U235 and U238 decays, you don't have to worry about the other daughter products. The second source of heat was from the chain reaction. This amount of heat is dictated by the reactor power. The reactor operation will dictate the power. I'm not sure why this question was closed, it asks about basic physics, not homework. | |
| Feb 26, 2022 at 20:28 | history | left closed in review |
Jon Custer ZeroTheHero John Rennie |
Original close reason(s) were not resolved | |
| S Feb 26, 2022 at 13:51 | review | Reopen votes | |||
| Feb 26, 2022 at 20:28 | |||||
| S Feb 26, 2022 at 13:51 | history | edited | Vogon Poet | CC BY-SA 4.0 |
Broad concept of calculating heat of radiation is asked, without any specific problem.
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| Feb 26, 2022 at 8:38 | history | closed |
StephenG - Help Ukraine John Rennie Jon Custer BioPhysicist jng224 |
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| Feb 25, 2022 at 14:24 | comment | added | Vogon Poet | @rmhleo That spawned my question. I thought there must have been some formula he used to predict how much graphite and uranium, and the diameter of the hole, etc., would generate $x$ joules of heat. CP-1 may have had the potential to go prompt critical, but it certainly was supercritical. | |
| Feb 25, 2022 at 14:20 | comment | added | Vogon Poet | @rmhleo CP-1 went supercritical and after achieving 200 Watts Fermi ordered it shut down. That's all they needed, and the Manhattan Project was authorized. The point is that there is no mention at all of some neutron source. He simply drilled holes in graphite and put rods of uranium in it, keeping cadmium poison control rods between each layer to absorb neutrons until ready. Maybe the neutron source is still classified? I do know that new cores use one, but I was told that was just for efficiency. | |
| Feb 25, 2022 at 13:28 | comment | added | rmhleo | No, there is heat indeed. And no neutron emitter was used. I think they used the neutrons produced in spontaneous fission of uranium isotopes 235 and 238. According to wikipedia, it operated at about 500 mW, which would be mostly dissipated into heat. | |
| Feb 23, 2022 at 16:43 | comment | added | Vogon Poet | Answer: No detectable heat at all without a neutron emitter? I missed that in Fermi's construction of CP-1. What did he use? | |
| Feb 23, 2022 at 14:58 | comment | added | Jon Custer | @VogonPoet - if you actually look at the U-238 decay chain (en.wikipedia.org/wiki/Uranium-238) you will notice that none of them are spontaneous fission. The nice decreases-by-4 mass units indicate alpha emission. U-238 requires energetic (>1MeV) neutrons to induce fission. | |
| Feb 23, 2022 at 13:21 | comment | added | Vogon Poet | Don't the trace amount of U-238 -> Th-234 -> Pa-234 -> U-234 -> Th-230 -> Ra-236 -> Rn-222 etc. have a potential to do this? That's at least 5 neutron sources. Well, the question is really about heat even if supercriticality is not achieved. Is there no heat at all? | |
| Feb 23, 2022 at 10:23 | comment | added | rmhleo | Do you mean his experiment on nuclear fission? Then it also needs to include a neutron source. | |
| Feb 23, 2022 at 10:23 | history | edited | John Rennie |
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| Feb 23, 2022 at 7:40 | review | Close votes | |||
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| Feb 23, 2022 at 7:15 | history | asked | Vogon Poet | CC BY-SA 4.0 |