Delujoč reaktor hladne fuzije kmalu v vsakem domu!

Tole ima pa smisel. Se začneš kar tresti, ko kdo podvomi v ideje, ki si jih posvojil.

Andrea Rossi
October 9th, 2011 at 6:29 AM

Dear Peter Roe:
The 99 Celsius water remained in the E-Cat at the moment in which we had to discharge it to inject cold water to cool down more rapidly the E-Cat to disassemble it and allow the ayyendants to look inside was about 20 liters.
The small difference of weight (the E-Cat had a weight slightly heavier than when it has been weighted before the test) is due to residual water. The stored energy in this amount of water is not significant, of course. Where the not-calculated-power-power produced by the reactor is relevant is the heating up to 660-80 celsius of the surface of the E-Cat, which is more than 5000 square cm. But we have chosen to allow a very conservative approach, to allow not biased People to understand well.

http://www.journal-of-nuclear-physics.c...

How can 30% of nickel in Rossi’s reactor be transmuted into copper?

Abstract
In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: how can 30% of nickel in Rossi’s reactor be transmuted into copper? “Everything should be made as simple as possible, but not simpler”, says a guy. I apologizes if I am too simplistic here.

Introduction
The interest on Andrea Rossi’s Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “this seems to be impossible because the produced copper isotopes rapidly decay into Ni”. But how it works?

How it works
Following Focardi Rossi [3]. a Ni58 nucleus produces a Copper nucleus according to the reaction

Ni58 + p → Cu59

Copper nucleus Cu59 decays with positron (e+) and neutrino (ν) emission in Ni59 nucleus according to

Cu59 → Ni59 + ν + e+

Then (e+) annichilates with (e-) in two gamma-rays

e- + e+ → γ + γ

Starting [3] from Ni58 which is the more abundant isotope, we can obtain as described in the two above processes Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.

Of course how can a proton p gets captured by the Ni58 nucleus? (and subsequent Ni59, Ni60, Ni61 and Ni62). Following Stremmenos [4]. a neutron-like particle, an electron proton pair, a mini-atom, a proton masked as a neutron, gets captured by the Ni58.

If the masked proton becomes a neutron the result is Ni59.
In order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.

The same process holds for all the subsequent transformations, until Cu63.
It remains to be understood the issue of the gamma radiation in the MeV range.

Numbers
I am an electronic engineer, so I need easy numbers in order to understand.
However “Everything should be made as simple as possible, but not simpler”, says a guy. Maybe I am too simple here.
Let’s calculate.

MeV for each Ni transformation
I read that 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.

Nickel consumption
According to many blogs in the Internet “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?
I follow a procedure outlined in [2].
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.

Conclusions
Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions.

http://www.journal-of-nuclear-physics.c...

Povztek je tak, da matematika ne zanika Rossijeve trditve o tem, da naj bi se lahko v 6 mesecih 30% niklja pretvorilo v baker. Toplota, ki je nastala, ustreza tej številki.

21:51, 9 October 2011 (PDT)

Please participate in the E-Cat Case Design Project
http://peswiki.com/index.php/OS:Cold_Fu...

Andrea Rossi, the inventor of the E-Cat (Energy Catalyzer) has asked for the help of the alternative energy community, in designing the case for his home heating product. He has asked for the community to submit concepts, so he can purchase the best idea from the individual who submitted it!

tale Andrea Rossi je res en medisjki človek

toliko časa bo propagiral zadevo

da bo na koncu še knjigo napisal in postal slaven

kot človek ki je pritegnil pozornost širše množice

Loocas je 9. okt 2011 ob 22:11 izjavil:

Conclusions
Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions.

http://www.journal-of-nuclear-physics.c...

Kaj od podčrtanega ne razumeš:

The fusion of lighter elements in stars releases energy (and the mass that always accompanies it). For example, in the fusion of four hydrogen nuclei to form helium, seven-tenths of 1 percent of the mass is carried away from the system in the form of kinetic energy or other forms of energy (such as electromagnetic radiation)[2] However, the production of elements heavier than iron absorbs energy.

Vir: Nuclear fusion @ Wikipedia

Poglej si še tale graf: File:Binding energy curve - common isotopes.svg @ Wikipedia

da je vložil kar nekaj dela in tudi denarja.

Če je sploh vložil svojega, ga je vložil v upanju, da bo iz 10 jurjev naredil milijon. Dopuščam tudi možnost, da je nategnil celo samega sebe in dejansko verjame, kar govori.

upam, da ce je nekaj na tem, da bo vajeti v roke prevzela res kakna firma, ki se spozna na svoj foh

Če bi bilo kaj na tem, je čisto vseeno, kdo to dela, ker taka stvar preprosto ne more propasti. Ampak kaj, ko ni nič na tem.

O.

Če bi delovalo, bi lahko prišlo v široko potrošnjo tudi neizpiljeno. In tudi kar se cene tiče, je dovolj najvišja možna cena, da se ti še splača kupit. Ti ne samo, da bi rad imel čudež, ti bi rad imel lep in najcenejši čudež. Daj se malo zberi in začni trezno razmišljati.

O.

bi ti to kupil, neglede na to, če bi mogla biti grda naprava zavita v neko aluminjasto folijo sred tvoje kuhinje... ropotat kot zmešana... cevi vsepovsod okoli...

Seveda bi kupil, takoj. Ne bi bil pa tako nor kot ti, in ne bi tega sredi kuhinje postavil, ampak bi dal lepo v klet.

čez 5 let pa bo na trg prišla manjša prkupna, tiha napravica

Tebi nekdo obljublja free energy (so to speak), ti bi pa rekel, ah ne, bom počakal, da bo v lepši škatli in še cenejša. In potem mi tak govori o trolanju

O.

energetik je 10. okt 2011 ob 11:25 izjavil:

Ampak tip je po mesecih izpostavljanja gama sevanju še vedno pri močeh...

Tud gama foton ima svoj ponos...

free? tisti sort to speak si zatakni nekam... ce sas 20,000 eur, valda ni free... v tistih letih pa ko se bi ti to porvnilo, bi pa prisle 100x boljse in cenejse naprave ven...

mah dej.. ne da se mi vec... cisto brezpredmetno... ce bo prav, ce ne bo pa tudi prav

@il:
To samo mi neverniki in placanci naftne in energetske industrije dvomimo.

Loocas je 9. okt 2011 ob 22:11 izjavil:

How can 30% of nickel in Rossi’s reactor be transmuted into copper?

Abstract
In the present article I would like to answer a question posed by L. Kowalsky in a recent paper: how can 30% of nickel in Rossi’s reactor be transmuted into copper? “Everything should be made as simple as possible, but not simpler”, says a guy. I apologizes if I am too simplistic here.

Introduction
The interest on Andrea Rossi’s Nickel-Hydrogen Cold Fusion technology is accelerating [1]. However, Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. Kowalski [2]. says that “this seems to be impossible because the produced copper isotopes rapidly decay into Ni”. But how it works?

How it works
Following Focardi Rossi [3]. a Ni58 nucleus produces a Copper nucleus according to the reaction

Ni58 + p → Cu59

Copper nucleus Cu59 decays with positron (e+) and neutrino (ν) emission in Ni59 nucleus according to

Cu59 → Ni59 + ν + e+

Then (e+) annichilates with (e-) in two gamma-rays

e- + e+ → γ + γ

Starting [3] from Ni58 which is the more abundant isotope, we can obtain as described in the two above processes Copper formation and its successive decay in Nickel, producing Ni59, Ni60, Ni61 and Ni62. Because Cu63, which can be formed starting by Ni62, is stable and does not decay in Ni63, the chain stops at Ni62 (i.e. Cu63). Each process means some MeV.

Of course how can a proton p gets captured by the Ni58 nucleus? (and subsequent Ni59, Ni60, Ni61 and Ni62). Following Stremmenos [4]. a neutron-like particle, an electron proton pair, a mini-atom, a proton masked as a neutron, gets captured by the Ni58.

If the masked proton becomes a neutron the result is Ni59.
In order to have Cu59 (increase of atomic number from 28 to 29) the electron (of the masked proton) gets ejected from the nucleus. The masked proton becomes a proton.

The same process holds for all the subsequent transformations, until Cu63.
It remains to be understood the issue of the gamma radiation in the MeV range.

Numbers
I am an electronic engineer, so I need easy numbers in order to understand.
However “Everything should be made as simple as possible, but not simpler”, says a guy. Maybe I am too simple here.
Let’s calculate.

MeV for each Ni transformation
I read that 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.

Nickel consumption
According to many blogs in the Internet “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?
I follow a procedure outlined in [2].
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.

Conclusions
Rossi says that about 30% of nickel was turned into copper, after 6 months of uninterrupted operation. At first glance this seems to agree with calculations based on simple assumptions.

http://www.journal-of-nuclear-physics.c...

loca, razumeš 1% tega kar piše tu?

in naj že eden poišče nuklearni periodni sistem elementov pls!!!!!!

5 nukleonov mu je kar 5u. mislim kakšen debil je ta človek NOOOOOOOoooooooooooo

a misliš da believrji vedo kaj je 5 nukleonov ali 5u

NE

zatorej so te cifre zanje čisto dobre da verjamejo in se slinijo nad pričakovanjem
ter obenem še kak euro primaknejo v pušico temu človeku

ni mi jasno tudi od kod mu kar nevtroni za Cu 63 iz Ni 58...
ampak to verjetno sploh ni pomembno
le kdo bi se sekiral zaradi tega... glavno da stvar dela!

samo trollanje

glede na to da ima nek človek zavezan žakelj in pravi da je notri zmaj

in da se prerekamo o tem

ja trolanje - totalno trolanje

glede na to da ima nek človek zavezan žakelj in pravi da je notri zmaj

metaforično seveda

od kod pridejo potem bakrovi atomi v E-Cat, če ne z fuzijo.

a fuzija je odgovor če ne veš kako
?

fuzija je mislim da še najmanj verjeten odgovor
vse kaj drugega prej bi bilo verjetno

ker elementi ne nastajajo v laboratoriju ampak pod strašanskimi pritiski in v strašansi vročini
znotraj zvezd ali tokamakov

če bo pa možno kdaj fuzijo zganjati v laboratoriju je pa seveda odprto

ampak ta rossi trenutno deluje prevarant

Torej ne veš kako nastanejo bakrovi atomi v njegovem E-Catu?

fuzija ni dokazan razlog

Par elementov nikel-baker je bil za kandidata za hladno fuzijo večkrat omenjen že v devetdesetih prejšnega stoletja.

če misliš na fleishman-pons zadevo...

Problem je bil v katalizatorju, ki naj bi ga Rossi pač našel

ima kak dokaz?

tu si lahko prebereš več o tem "žaklju z zmajem"
Energy Catalyzer @ Wikipedia

naj poudarim:
According to PhysOrg, the demonstrations held from January to April 2011 had several flaws that make them lose credibility and Rossi has refused to perform several tests that could clarify dark points

Imaš novega starega nekaj dni z občutno več dokaznega materiala.

daj link

...
potem si pa preberi najnovejše na wikipedii
link sem ti že prilimal

je pa zanimivo da pol leta človek še vedno skriva zadevo kot kača noge

In August 2011 Rossi announced that he had cancelled his contract with Defkalion, stating that none of E-Cat's technology had been transferred to them.[49][50][51][52][53] Rossi said that "the issue [that caused the cancellation] is just financial, not personal, nor technological, nor scientific", and that he wouldn't comment further because he was preparing a lawsuit against Defkalion.[49] A board member of Defkalion likewise indicated that the reasons for the cancellation were entirely financial.[54] Rossi later confirmed this, specifying that Defkalion was unable to fulfil its contractual obligations in due time.

in pa še iz tega meseca

In early October, Rossi stated however that due to contractual difficulties, the container had not been shipped to the U.S.

torej.. niti enega linka nisi odprl, ki sem jih nalimal tukaj na zadnjih 2-3 straneh?

prilimaj link od wikipedie ali od kake resne strani

razni tvoji reallyrealscience linki me ne zanimajo

Dear Don Witcher:

no pa je našel nekoga

Tudi če zadeva zlavfa je cena 200€ za tistih 5kWh vseeno malce pretirana :)

Aja: Kam potem z vsemi nuklearnimi odpadki ? Menda ne kar na deponijo ?

- Še vedno ostaja vprašanje, kaj čaka s svojo 1MW elektrarno. Mar jo nebi poizkusil že zagnati do sedaj - vsaj poizkusno?

In zakaj niso delal benchmarkov na tisti elektrarni, saj je vse že pripravljeno.
+ Zakaj je pri testu bilo vse skupaj povito z alu folijo ? Če je zadeva tako energetsko pozitivna se ti lahko je*e za izgube ....