More on Rossi / Focardi LENR / "Cold Fusion"
Andrea Rossi is not wasting time. As noted here earlier, Rossi has invested almost all his money in developing and producing the LENR reactors for the Athens, Greece 1 MW plant. Rossi also claims to have signed a contract "of tremendous importance" in the USA, with a company he is not yet at liberty to name.
Rossi plans to install a one megawatt, American made E-Cat power station in a factory in Greece in October, 2011. Rossi believes that only a working commercial power station can definitively prove to the world that his creation is real. If E-Cats turn out to be as economical as expected, they will eventually be used to power cars, trucks, trains, ships, aircraft, and spacecraft. _OpEdNews Chris CalderMore background and several helpful links at above link.
Update: More from Next Big Future
Both Brian Wang and Brian Westenhaus have been following the progress of the Rossi / Focardi low energy nuclear reaction device.
Rossi claims that the reactor is able to obtain large amounts of heat energy from the low energy nuclear transmutation reactions that transform Nickel into Copper. Here is a more detailed look at the energy numbers involved in such a transmutation:MeV for each Ni transformation
Starting from Ni58 we can obtain Copper formation and its successive decay in Nickel, producing Ni59, Ni60, and Ni62. The chain stops at Cu63 stable.
For simplicity I assume all the Nickel in the reactor in the form Ni58.
For simplicity I suppose for each Ni58 the whole sequence of events from Ni58 to Cu63 and as a rough estimate I calculate the mass defect between (Ni58 plus 5 nucleons) and the final state Cu63.
Ni58 mass is calculated to be 57.95380± 15 amu
The actual mass of a copper-Cu63 nucleus is 62.91367 amu
Mass of Ni58 plus 5 nucleons is 57.95380+5=62.95380 amu
Mass defect is 62.95380-62.91367=0.04013 amu
1 amu = 931 MeV is used as a standard conversion
0.04013×931 MeV=37.36 MeV
So each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.
One hundred grams of nickel powder can power a 10 kW unit for a minimum of six months.
How much of Ni58 should be transformed, in six months of continuous operation, in order to generate 10 kW?
10 kW is thermal or electrical power. The nuclear power must be larger. Assume a nuclear power twice:
20 kW = 20,000 J/s = 1.25 x 10**17 MeV/s.
Each transformation of Ni58 into Cu63 releases 37.36MeV of nuclear energy.
The number of Ni58 transformations should thus be equal to (1.25 x 10**17)/37.36 = 3.346 x 10**15 per second.
Multiplying by the number of seconds in six months (1.55 x 10**7) the total number of transformed Ni58 nuclei is 5.186 x 10**22.
This means 5 grams.
The order of magnitude is not exactly the same but seems to be plausible. This means also 5 grams of Nickel in Rossi’s reactor transmuted into (stable) Copper after six months of continuous operation at the rate of 10 kW. _NextBigFuture
This may seem incredible to most persons who know how many tons of coal are required to provide the same amount of power as 5 grams of nickel. But nuclear energy is on a far different level of scale than chemical energies, such as combustion energy.
But if you consult this table of energy densities provided at Transtronics Wiki, you can clearly see the difference in scale between the energy of nuclear reactions and the energy from chemical reactions -- roughly 7 or 8 orders of magnitude, depending on the method of comparison.
Imagine the savings in fuel transportation costs alone!
Will this sparkling new form of energy prove to be true gold, or just a fool's flash in the pan? Time will tell. _AlFin2100