KitemanSA wrote:stefanbanev wrote:Dan Tibbets> You could cycle between say nickel and uranium,
Dan Tibbets> gaining energy from both pathways.
Yes, apparently such E+ cycle may be drafted:
i) Ni+p...Cu+p ...Pb+p..........-> U235 (fission-bang) go to (i)
Every step decreases the mass according M -= Ei/CC
Note(*): the fission products should have enough light enough atoms to ensure exothermic path.
Dan Tibbets> You now have an inexhaustible free energy system.
Absolutely not, once mass == 0 thee is no E.
Needless to say that such cycle is highly hypothetical nevertheless, it shows that 100% mass->energy conversion by means of nuclear reaction may be possible if Note(*) can be provided.
Stefan
As I pointed out in another post, the supply of energy is only as "inexhaustible" as the supply of protons and free neutrons. Once you no longer have either of those, the process stops. Not sure why he refuses to get that point.
Actually for nucleosythesis of intermediate and heavy nuclei, alpha particals are the best ingredient from a stability standpoint.
You realize that you are ignoring many links to physics and stellar sites and texts that state repeatedly and unequivocally that the binding energy per nucleon determines the energy flow direction. It is not the ingredients so much as the end product . Nickel is the end of the line. The the holy grail, the most stable nucleus, the most tightly packed nucleus, the nucleus with the highest binding energy per nucleon.
Keep in mind that the proton has no binding energy. Any number that might be quoted is based on the weight of the proton in the stable carbon 12 nucleus. This is by convention, not on any underlieing property of the free proton. . Yet you say that the protons binding energy is added to the product nucleus. Where does this energy come from? It doesn't come from anywhere. You are only shifting the mass energy proportions within the product nucleus. Otherwise you would be creating energy out of nothing- a free energy, perpetual motion machine. Again, and again, and yet again, it is the density/ packing of the nucleons, specifically the average nucleon that is important. It is not a uniform homogenous environment within the nucleus. This is the property that determines the endothermic vs exothermic reaction.
You say that so long as there is hydrogen (protons) present, atoms can be built to unlimited size. You have no other limit in you system. This despite obvious consequences that is counter to the limits on how heavy a nucleus can become, and what drives radioactive decay and fission. I have pointed you to numerous sites that clearly defines the pertinent predictors and the reasoning behind them.
Finally. The total binding energy goes up indefinatly based on the graph you insist on using, so if you use this as the only factor there is absolutely no reason why the fusion of any combination of nuclei will not release energy and become more stable, after all- the total binding energy is you predictor. You could merge two uranium nuclei , get energy out, and have a more stable nucleus. A star would condense into one single huge nucleus, before it ran out of energy to resist gravitational collapse (I an talking about nuclear fusion here, not other effects like overcoming the Pauli exclusion principle) or matter- antimatter annihilation. Using the excuse that free protons are needed is immaterial to this prediction, alphas, iron nuclei, Californium nuclei, etc would all work, the end product would always have more total binding energy. All of the physicists and astronomers are obviously stupid for presenting the binding energy per nucleon curve, with arrows and discussions on how Ni62 is the peak (or minimum if the reciprocal of the graph is used as some prefer), and turn around point . Your interpretation also has no room for exothermic heavy element fission reactions.
IF both both fusion and fission can produce- release excess energy, there has to be a tipping point or minimal energy (most stable) state between them, otherwise, again, it is a perpetual motion machine.
Basically you are confused about the energy contained within a nucleus and the energy that is released or absorbed in the formation of a nucleus through fusion/ fission mechanisms. Your arguments would be valid if you were talking about the total annihilation of a nucleus , but these nuclear processes are dealing with only a tiny and variable fraction of this total energy (~0.5% of the annihilation energy), whether it is in the form of binding energy, mass deficit, rest mass of the particles, etc.
Dan Tibbets
To error is human... and I'm very human.
