26 ms·

Perhaps I fail to understand, but doesn't this comparison depend on a number of parameters such as the total reactor fuel load and enrichment, the burn rate, th

Perhaps I fail to understand, but doesn't this comparison depend on a number of parameters such as the total reactor fuel load and enrichment, the burn rate, the cost of nuclear fuel, the cost of solar PV, the lifetimes of each system, and the relative process efficiencies (notably the cost of decommissioning nuclear)? Otherwise you might as well say a teaspoon (or whatever) of water has as much potential fusion energy as 1 Kg U235 at a fraction of the price. ;-)

internetter 3y ago
Yes, the amount of estimations I made to get to that number is absurd, and very much "best case" with no regard for inefficiencies (both nuclear and solar systems are currently leaving lots on the table).
credit_guy 3y ago
Small nitpick: one teaspoon of water has much less potential fusion energy than 1 kg of U235, and actually much much less than 1g of U235, even allowing for fusion technology that does not exist and will not exist in 50 years. Here's why. The Sun transforms hydrogen into helium. But that's a fairly complex chain and nobody in the industry or academia is trying to replicate that. When people talk about fusion, here's [1] the reactions they are considering. The best yielding fusion reaction is deuterium-tritium and deuterium-helium3 [1]. Tritium and helium-3 virtually don't occur naturally on Earth, and deuterium is very rare, at about 0.02% of the hydrogen. A teaspoon of water contains about 0.5 grams of hydrogen, and out of that about 0.0001 grams of deuterium. Let's say that someone magically brings the necessary tritium or helium-3. How does that compare with 1 gram of U235? The fission of 1 nucleus of U235 yields about 190 MeV of energy. 1 MeV is one megaelectronvolt, and is a unit of energy. It does not matter how it translates into joules or watt-hours. It is the unit used when talking about fission and fusion. So, 235 nucleons produce 190 MeV, which is about 0.8 MeV per nucleon. The two reactions mentioned involve 5 nucleons and yield about 18 MeV, which means 3.6 MeV per nucleon or 4.5 times more per nucleon than U235. So, even if all the hydrogen in the one teaspoon of water was Deuterium and Tritium, in the correct ratios to do the fusion, we'd get only 4.5 times more energy than from one gram of U235. In reality, from one teaspoon of water we'd extract a very tiny amount of deuterium that's usable, and we'd need to breed Tritium or Helium-3 separately. By the way, separating deuterium from water is a very expensive process. The Nazis tried to do it during WW2, and they were doing it in Norway. Once the British special forces destroyed the plant, the Nazis could not restart the heavy water production, and their atomic project basically stopped then and there. [1] https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_ca...

internetter 3y ago

Yes, the amount of estimations I made to get to that number is absurd, and very much "best case" with no regard for inefficiencies (both nuclear and solar systems are currently leaving lots on the table).

credit_guy 3y ago

Small nitpick: one teaspoon of water has much less potential fusion energy than 1 kg of U235, and actually much much less than 1g of U235, even allowing for fusion technology that does not exist and will not exist in 50 years. Here's why. The Sun transforms hydrogen into helium. But that's a fairly complex chain and nobody in the industry or academia is trying to replicate that. When people talk about fusion, here's [1] the reactions they are considering. The best yielding fusion reaction is deuterium-tritium and deuterium-helium3 [1]. Tritium and helium-3 virtually don't occur naturally on Earth, and deuterium is very rare, at about 0.02% of the hydrogen. A teaspoon of water contains about 0.5 grams of hydrogen, and out of that about 0.0001 grams of deuterium. Let's say that someone magically brings the necessary tritium or helium-3. How does that compare with 1 gram of U235? The fission of 1 nucleus of U235 yields about 190 MeV of energy. 1 MeV is one megaelectronvolt, and is a unit of energy. It does not matter how it translates into joules or watt-hours. It is the unit used when talking about fission and fusion. So, 235 nucleons produce 190 MeV, which is about 0.8 MeV per nucleon. The two reactions mentioned involve 5 nucleons and yield about 18 MeV, which means 3.6 MeV per nucleon or 4.5 times more per nucleon than U235. So, even if all the hydrogen in the one teaspoon of water was Deuterium and Tritium, in the correct ratios to do the fusion, we'd get only 4.5 times more energy than from one gram of U235. In reality, from one teaspoon of water we'd extract a very tiny amount of deuterium that's usable, and we'd need to breed Tritium or Helium-3 separately. By the way, separating deuterium from water is a very expensive process. The Nazis tried to do it during WW2, and they were doing it in Norway. Once the British special forces destroyed the plant, the Nazis could not restart the heavy water production, and their atomic project basically stopped then and there. [1] https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_candidates_for_terrestrial_reactions https://en.wikipedia.org/wiki/Nuclear_fusion#Criteria_and_ca...