by
Wladimir Guglinski
retired, author of the Quantum Ring Theory
.
.
Abstract
Dr. Wilfried Nörtershäuser of the Helmhotz Center for Heavy Ion Research at the University in Mainz says on the 2009 experiment which had detected a neutron halo in 4Be11 with distance 7fm from the cluster:
“By studing neutron halos, scientists hope to gain further understanding of the forces within the atomic nucleus that bind atoms together, taking into account the fact that the degree of displacement of halo neutrons from the atomic nuclear core is incompatible with the concepts of classical nuclear physics”[ 2 ]
In the case of 4Be11, the halo neutron and the nuclear core are separated by the distance of 7fm, and so such isotope represents the experimental proof that the cohesion of nucleons within the light isotopes cannot be promoted by the strong nuclear force.
Such experimental discovery published in 2009 had been predicted years ago, because according to the new nuclear model proposed in Quantum Ring Theory, published in 2006, the cohesion of the nucleons within the light nuclei is not caused by the strong nuclear force.
Here in this paper the new nuclear model is submitted to a scrutinity so that to verify whether from its structure it’s possible to explain the stability of the light nuclei and to reproduce the nuclear properties as nuclear spins, electric quadrupole moments, and magnetic moments. Nuclear magnetic moments are calculated from two different and independent methods. In the second, named “method of equilibrium between nucleons”, it’s presented the Lagrangian for nuclei with Z < 8. The results obtained from them agree each other, and are corroborated by nuclear spins and electric quadrupole moments suplied by nuclear tables.
In this Part One are presented calculations on magnetic moments for the isotopes of lithium, beryllium, and boron. In the next paper Part Two will be exhibited calculations for carbon, nitrogen, and oxygen. In the paper Part Three the author will exhibit calculations for electric quadrupole moments.
.
.
Hello Dr Rossi, when he thinks that its technology will spread quickly? when he expects there will be a boom in sales and orders? has been making predictions?
Dear Dr Joseph Fine:
Thank you very much, we will study your proposal.
Warm Regards,
A.R.
To the readers of JONP, This information is given as a follow-up to my previous part two information that may help better explain the subject of aether with regards quark structure and its internal dynamics which although forms the basic building blocks of the material world is of different internal dynamics than that of structure which quarks form of the atomic dimensions.
To re-cap, aether has no charge because it contains no gravity i.e. it is mass less. To visualize a triplicity of aether in the structure of a quark (that is the result of the creating environment as previously mentioned) simply draw a horizontal line and on the line draw draw three capital Ds, each progressively larger than the next, right to left. These three Ds rotate upon the diametrical dimension of the small inner D. This createsa top and bottom well. Top well, strong gravity value (north pole). Bottom well, weak gravity value (south pole). No internal gravity between poles. These polar values are with regards the creating environmental influence with regards position. These Ds are of aether routes and originate from one radial dimension of the creating environment as described previously. Thereby consider that these Ds are on a slight angle, large D top of angle. Starting at the top of the small D an aether travels on its circumference down to the base of its D and almost at the same instant the middle Ds aether travels down on its circumference to the base of its D. Then almost at the same instant the large Ds aether travels down on its circumference to the base of its D and is drawn into the well. It is because of the timing sequence of these events that the quark can represent a kinetic neutral energy knot. The small inner D has the most positive gravity with regards environmental position, being at the most depth/inner position of. The aether of the small D senses the descent of the aether of the middle D and thereby travels to the base of the large D whereupon it ascends to its top on its circumference. Where does the aether of the middle D travel?. The two aethers of the small and large D cycle round in unison in an alternating fashion, one always on the circumference of the large D and one always on the diameter of the small D. The aether on the small D always travels on a helical trajectory of the diametrical dimension but when it travels the circumferential dimension its trajectory becomes more linear and its velocity increases so as to keep sync with its other aether partner which does likewise. Thereby with this configuration and timing sequence the aether of the middle D plays an important role as it alternates between the diametrical and circumferential dimension on a radial to occupy vacant space exactly between both alternating aethers of the structure. As previously explained only 180 degrees with a rotation is required to achieve 360 degrees of structure. Aether velocity is the key to solidity with regards quark structure. Environmental position thereby is responsible for the density of substance. Only two opposite dimensions exist. Quarks can be considered kinetic neutral energy knots of no definite size and consequently they can occupy various positions with regards environmental conditions, thereby potentiality could be said to be dependent upon environmental factors that dictate a quarks size. Also because quarks are formed on the internal radial dimensions of the creating environment, there ultimate destruction is only achieved on the exterior radial dimension of the creating environment these being created by exterior gravitational forces encountered.
This information could be worth considering, especially when these initial building blocks are put into the atomic dimensions that I will put forward at a later date as considerations. By understanding quarks and there role in the atomic dimensions I do believe it will help, in part, to the understanding of LENR. In the meantime Best regards, Eric Ashworth.
Andrea Rossi,
I did a back-of-the napkin calculation of one possible arrangement of 108 (10-KW) cylinders in such a way as to fit into an enclosed volume of (approx.) 3 cubic meters. The shape of the container in this example, is a tube with a 36-side polygonal cross-section and a height of 3 meters.
The vertically oriented 36-gon, has 3 layers of 1 meter height each, with each layer containing 36 cylindrical modules. If you ignore the internal volume of the tube, considering only the enclosed modules, each layer encloses approx. 1 cubic meter (for 36 modules), so each module uses a volume of approx. 1/36 cubic meter.
This seems to work out if the module radius is about 9-10 cm and 1 meter high and the container radius is at least 1.25 meters. So the modules are between the inner and outer wall of the container.
The separation between the inner and outer walls of the container are roughly 20-25 cm in thickness (about 10 inches).
As usual, this is only my idea, but it does allow a person to stand up inside the container, while the enclosed module volume can still be claimed to be 3 cubic meters. (Total volume is still more than 3 cubic meters.)
As usual, your actual results may be different.
Geometrical regards,
Joseph Fine
Dear Andrea,
According to the activity on your blog, it seems that you have -finally- found some time to rest after the delivery of the plants.
To Steven N Karels, you answered that you also delivered experimental protoypes to your customer.
Did you also deliver the newer and experimental mouse-and-cat version in the 1kW-range version ?
I thought this one to be important for electricity generation using a Carnot-cycle, and I was wondering if the cooperation with the customer also tends towards the building of electrical power generation plants. I cannot think of other reasons to deliver experimental devices.
Kind Regards
Koen
Dear Steven N. Karels:
Sorry for my misunderstanding.
The maintainance is made by our certified operators going inside the container of the plant. Substantially, we substitute the E-Cats modules, bring home the old ones to recycle the charge, and check all the set up, even the electronics.
Warm Regards,
A.R.
Dear Andrea Rossi,
I think there is a mis-understanding. I use the term “Thermal eCat” to refer to the 110degC unit. I use the term “Hot eCat” to refer to the unit that can output 600 degC temperatures.
So I think whe can agree that you delivered a Thermal eCat under my description above.
If you have a preferred nomenclature for describing the two classes of eCats, please state them and I will comply in the future.
The 106 modules is essentially the same as my comment of “there are 100 or maybe more units per 1MW unit”
I don’t see how the specification of the input voltage is confidential. In the US, the standard industrial voltages are nominally 220VAC and 440VAC. The amount of current is a function of that voltage, the COP (you stated 6 or more) and the output power (you stated 1MW). I had posted some months ago the same voltage and current statement and you agreed to them as being correct.
Since you replied that my understanding was incorrect on maintenance, can you please clarify how maintenance is intended to be accomplished for the 1MW Thermal eCat unit?
Dr. Rossi, please Google:
$1 billion transmission line.
Click on:
Buinessweek
I was having some fun with these numbers, but I don’t have all the numbers.
You may or may not want to play my game.
It will take 3 years and 5 billion dollars to build a 6,000
MW Dam.
Amount of loan for Transmission Line, $1 billion.
Cost of Line $1.2 billion.
Length of line 385 miles.
Transmission Line to move electricity from Dam to City.
Cost of 6,000 MW:
Dam, $5 billion
E-Cat, ?
Amount of Land:
Dam, ?
E-Cat, ?
Cost of Transmission Line:
Dam, $1.2 billion
E-Cat, Zero
Electricity produced during 3 years of construction, Zero.
During that same 3 years, E-Cat
will be producing 6.000 MW per year.
Cost of Fuel per year:
Dam, Zero
E-Cat, ?
Cost of maintenance per year: Dam. ?
E-Cat. ?
Cost of shipping:
Dam, Zero
E-Cat, ?
So maybe the Dam is less espensive ?
The way I understand one billion.
If you had one pile of one million dollars, it would take one thousand piles of one million dollars in each pile to equal one billion dollars.
Robert Curto
Ft Lauderdale, Florida
USA
Dear Steven N. Karels:
0- I did not say that we delivered a 1 MW Hot Cat, I said that a 1 MW Hot Cat should have that volume, but we did not yed make one. We made smaller modules, still in a phase of industrialization. What we delivered is a low temperature 1 MW E-Cat plus experimental prototypes of Hot Cat, fueled by electric power or gas.
1- No
2- The 1 MW low temperature plant has 106 modules, each of 10 kW
3- This information is confidential
Warm Regards,
A.R.
Dear Andrea Rossi,
Interpreting your response to Dr. Joseph Fine —
The delivered 1MW Thermal eCat has a volume of 3 cubic meters or smaller. Therefore, one does not walk into it for maintenance but one disassembles it (while it is not operating) for maintenance purposes. The basic Thermal eCat module is a 10kW producing module, Therefore, there are 100 or maybe more units per 1MW unit. From this “matrix” of 10kW eCats, a maintenance person would remove and replace a single 10kW eCat module as the “field replaceable unit” of the 1MW eCat.
Additional questions:
1. Is the above essentially correct?
2. Is the 1MW unit one unit or two units (i.e., control unit and power production unit)?
3. Assuming a COP of 6 (or higher) means the electrical load will be 167kW for a thermal output of 1MW. Assuming it runs off of 220VAC power, the current will be about 760 Amps. Is this the actual operating voltage and load?
Dear Dr Joseph Fine:
0- the photos of the delivery test will be published on this blog within several days, after the permission of the Customer
1- internally
2- right
3- the quantum of a 1 MW plant is 10 kW
Warm Regards,
A.R.
Wladimir,
How does QRT explain the electric quadrupole moment of zero for 2He4? (Does it involve treating 2He4 like two protons and two neutrons, with the two protons performing a sphere? Or is it necessary to deconstruct 2He4 further into four protons and two electrons, with the four protons performing a sphere? In the latter case, how would the charge of the two electrons affect the electric quadrupole moment?)
All the best,
Joe
Andrea,
Greetings,
Soon, I hope you will be able to present some pictures of the recently delivered E-Cat and Hot-Cat systems. Until the Hot-Cat is presented, can you give me/(us) some idea of the configuration of the Hot-Cat container. The E-Cat is in a shipping container that is, basically, a large box (or rectangular parallelepiped).
I think the Hot Cat container may be cylindrical in shape and may be divided in several sections in order to gain access to each of the component modules.
1) How does maintenance access the component modules? By walking inside the container (when it is shut down) or by opening access panels from outside of the cylinder (or box)?
2) I forgot the volume of the Hot-Cat system. I recall that it is significantly smaller than the E-Cat container. I remember a volume of about 3 cubic meters or less. Is that almost right?
3) Is the (1 MW) Hot Cat already divided into two or more connected subsections or if not, can it be divided into two smaller sections, essentially making it into a pair of 1/2 MW Hot-Cats.
Best wishes for a wonderful summer,
Joseph Fine
Dear Steven N. Karels:
1- more than 6
2- continuous
Warm Regards,
A.R.
Dear Mark Saker:
1- this technology is an unseparable part of me
2- no, not time for them if I want to make this well
3- I wouldn’t say so
4- it will be done soon
Warm Regards,
A.R.
Dear Gian Luca:
I think it will be before, yes.
Warm Regards,
A.R.
Joe wrote in May 1st, 2013 at 4:11 AM
Wladimir,
1. You stated, “Yes, the magnetic moment induced by the magnetic field of 2He4 is NULL, since its TOTAL magnetic moment is zero.” (April 30th, 2013 at 6:09 AM) You mention a magnetic field for 2He4, but would not a magnetic moment of zero indicate a LACK of magnetic field?
RESPONSE:
A) Total null magnetic moment due to the two deuterons because one deuteron has µ=+0,857 and the other has µ=-0,857
B) Total null magnetic induced by the rotation of the charges of the two protons (because of my explanation in April 30th, 2013 at 9:27 PM)
.
2. Could not standard physics describe 2He4 as two deuterons with antiparallel spins and orbiting a barycentre with opposing angular velocities?
RESPONSE:
A) two deuterons with antiparallel spins and angular velocity in the same direction have total µ= -0,857+0,857 =0 , and total spin i=0.
But the two charges 1,6×10^-19C of the protons induce a total magnetic moment different of zero, because they gyrate in the same direction.
B) two deuterons with antiparallel spins and opposing angular velocity have total µ=+0,857+0,857 = + 1,714
Regards
wlad
Thank you A.
I hope that will be before 6 months.
In the meantime we can give you some “medicine” for
fool anxiety and reassure us?
Buon lavoro a Lei e tutta la sua equipe!
Dear Andrea,
Congratulations on making the delivery date, I’m sure it was a relief. I have a couple of questions please:
1. Do you think the rest of your working life will be related to the ecat, or do you plan to hand over R&D to the partner? Are there other projects you are interested in that you would like to work on?
3. Has the refill process changed dramatically for the newest 1kw reactors?
4. Pictures picture pictures, perhaps just remove anything that might identify shipping so you can publish now 🙂
Many thanks
Mark
Dear Andrea Rossi,
oh, strange… try this alternative link for “periodic table of metals”
http://d24w6bsrhbeh9d.cloudfront.net/photo/377722_700b.jpg
or google “periodic table of metals” and click the black one
Sound regards
Silvio
Dear Andrea Rossi,
Can you share with us what was the average COP observed during the acceptance testing?
Was the 24 hours of testing continuous or was it divided over several days?
Dear Gian Luca:
Oh, I understand. The Indipendent Third Party will surely publish the report of the test made. I have to repeat that the “when and where” does not depend on me. Consider that the time that normally is spent between a test made by a third party and the publication is 6 months. The tests have been completed on the 24th of March, but I think the publication will arrive before 6 months will be expired from that date.
Warm Regards,
A.R.
julian_becker:
we will publish the photos our Customer will allow us to publish, after the plant will have been arrived at destination.
Warm Regards,
A.R.
Dear Steven N. Karels:
0- yes
1- the plant has been tested to check COP and safety.
2- 24 hours
3- of course
4- will come now
5- some
6- in this case our Customer does all
Warm Regards,
A.R.
Dear Silvio Caggia:
sorry, the link doesn’t open anything to me.
Warm Regards,
A.R.
Dear Bernie:
For security reasons we prefer not to give information about the shipping.
Warm Regards,
A.R.
Dr. Rossi: Congratulations on the shipment. I assume the buyer is responsible for shipping to US, is that correct?
Dear Andrea Rossi,
Given the “Periodic table of metals” linked below, can you point your finger on “elements” that helped you to invent the e-cat?
🙂
Sound regards
Silvio
images3.wikia.nocookie.net/__cb20090815210458/nonciclopedia/images/c/c2/Tavola_periodica_dei_metalli.jpg
Dear Andrea Rossi,
Congratulations of the delivery of the systems. I understand “delivery” to mean acceptance of the systems by the Customer at your location.
1. Any chance of releasing the nature and pass/fail conditions of the acceptance tests?
2. How long did the acceptance testing last (hours, days?)
3. Can I assume the Customer had his technical contract representatives at your location to witness the testing?
4. Was technical documentation (training documentation, users manual, etc) provided at the time of acceptance or will that come later?
5. Were spare parts included with the order? If so, what is the maintenance philosophy (your people do all the maintenance on-call or is the Customer having his maintenance personnel trained)?
dear mr. Rossi, when can we see the images. Eagerly awaiting them. Can you just publish a few now? People are going crazy about waiting for them I believe. Best regards, Julian
excuse me A.R.
“T.P.” …… report of the Third Party.
Dear Neri B.:
The delivery, after an acceptance test, has been made today.
Dear Neri B.:
The delivery has been made today.
the photos of the plant will surely be published.
We will publish them on the Journal Of Nuclear Physics
Warm Regards,
A.R.
Dear Gherardo:
You will find some.
Warm Regards,
A.R.
Dear Gian Luca:
What the heck is T.P. ?
Thanks a lot to you,
A.R.
Dear Andrea,
Could you please update us on the plant delivery? There were some delays or everything went good? I am also courious about the pictures of the plant and the “2 stage” module..can you tell us when and where you are going to publish them?
Thank you. Let the revolution begin
Neri B.
Carissimo A.R.
there are a news about T.P.?
Thanks a lot.
Looking for pictures…
Wladimir,
1. You stated, “Yes, the magnetic moment induced by the magnetic field of 2He4 is NULL, since its TOTAL magnetic moment is zero.” (April 30th, 2013 at 6:09 AM) You mention a magnetic field for 2He4, but would not a magnetic moment of zero indicate a LACK of magnetic field?
2. Could not standard physics describe 2He4 as two deuterons with antiparallel spins and orbiting a barycentre with opposing angular velocities?
All the best,
Joe
Dear Joe
I was thinking about the rotation of the 2He4 in the 5B10, and had concluded the following:
The rotation of the charge of the 2He4 does not induce magnetic moment.
I will explain why.
1- Look at the Fig. 1 in the page 4, showing the fluxes n(o)-up and n(o)-down crossing the two protons of the 2He4.
2- Look at the Fig. 2 in the page 5, showing the structure of the 8O16 nucleus. In its structure we have the following:
a) The proton of the deuteron D-6 is crossed by a flux n(o)-up, and the proton of the deuteron D-3 is crossed by a flux n(o)-down.
b) The sign of the magnetic moment induced by the charge of the proton of D-6 depends on the direction of the flux n(o)-up.
c) The sign of the magnetic moment induced by the charge of the proton of D-3 depends on the direction of the flux n(o)-down.
d) Therefore the sign of the magnetic moments induded by the charges of the two protons in D-6 and D-3 are contrary, one is positive, and the other is negative. So, thanks to the contribution of the fluxes n(o)-up and n(o)-down, it is null the total the magnetic moments due to the charges of the two deuterons situated symmetrically with regard to the z-axis.
No let’s analyse the nucleus 2He4 shown in the Fig. 1.
One of the two protons of the 2He4 is crossed by n(o)-up, and the other is crossed by n(o)-down.
Therefore the total magnetic moment due to the rotation of the two charges in the 2He4 is null.
IT’S IMPOSSIBLE TO EXPLAIN THE NUCLEAR PROPERTIES OF THE NUCLEI WITH Z=N=PAIR, by considering the nuclear models of Classic Nuclear Physics.
Dear Joe,
there is no way to explain why nuclei with Z=N=pair have magnetic moment zero together with null quadrupole electric moment, by considering the Standard Nuclear Physics.
Let me explain why.
Consider the nucleus 2He4.
The experiments show that it has magnetic moment zero and null nuclear spin, and null quadrupole electric moment.
According to the Standard Nuclear Physics, in order to explain the null magnetic moment of 2He4 we have to consider the following two hypothesis:
1- First hypothesis
The two protons move with antiparallel spins, toward the same direction. So, the magnetic moment due to the magnetic moment of each proton is:
µ= −2,973 + 2,973 = 0
Their total spin is i= +1/2 – 1/2 = 0
However, as the two protons move toward the same direction, their two charges induce two magnetic moments with the same sign. Let’s consider that each charge of the protons induces a magnetic moment µ= A, due to their rotation.
The total magnetic moment induced by the two charges is:
µ= +A + A = 2A.
Therefore the two protons cannot move with antiparallel spins toward the same direction, because the magnetic moment of 2He4 would not be zero.
2-Second hypothesis
The two protons move with parallel spins, toward contrary directions (one moves clockwise, and the other counter clockwise). So, the magnetic moment due to the magnetic moment of each proton is:
µ= −2,973 + 2,973 = 0
The magnetic moment induced by the charge of the two protons is zero, because they move in contrary directions.
However the total spin of 2He4 cannot be zero, since the two protons have paralell spins, i = 1/2 +1/2 = +1
In order to get a null total spin of 2He4 the two neutrons have to move with parallel spins, in contrary direction:
i= -1/2 – 1/2 = -1
As they move in contrary direction, the total magnetic moment of the two neutrons is null.
Therefore we have:
One proton and one neutron with clockwise orbit
One proton and one neutron with counter clockwise orbit.
But such structure of the hypothesis 2 cannot have chaotic spin. As the four nucleons gyrate about the same axis, they form a disk with a big inertia moment.
Then the experiments cannot detect quadrupole electric moment zero, because the two protons do not form a spherical shape.
The experiments can detect zero quadrupole electric moment only in the case when the 2He4 has chaotic spin. But with the structure of the hypotesis 2 it is impossible to have a chaotic spin.
Therefore by considering both the hypothesis 1 and 2 it’s IMPOSSIBLE to explain the nuclear properties of 2He4, by considering the Standard Nuclear Physics
The same considerations we can apply to any nucleus witb Z=N=pair.
Regards
wlad
Dear Paul:
1- depends on the energy involved in vibrations or stress
2- in future maybe
3- no
4- before talking of this issue the safety certification issue must be resolved. Theoretically yes.
5- not possible for now, but with a massive production the prices will drop.
6- I have no idea
Thank you very much for your kind feeling,
Warm Regards,
A.R.
Dear Dr. Rossi,
I have been following your progress since your demonstration in Jan 2011, and periodically monitoring cold fusion since 1989. I have a few questions that I cannot recall being answered before; at least not in quite this form.
1. Is the reaction disturbed or degraded if the apparatus is moved or vibrated? I am particularly interested to know if it could be adapted to operate in a vehicle such as a car, or truck, or ship, or even an aeroplane.
2. Do you foresee that it could be installed (with suitable batteries and electrical generators) to deliver power as a stand-alone unit?
3. Can you estimate the expected power to weight ratio for such an arrangement?
4. Do you consider that it will be possible to construct houses with the e-Cat as the sole power source? For instance a house constructed in a wilderness area without electrical grid access?
5. You have provided anticipated cost estimates at various times in the past. Can you guess at the cost of a unit as defined in 2 above delivering 10 kw electric? Please guess both the initial capital investment and the running costs per year and per kwh.
6. Could an arrangement as described in 2 be operated in space, or is there something in the e-Cat that specifically requires gravity?
I am a student of economics and I am particularly interested in these details in order to estimate the possible economic impacts.
I am reasonably convinced that the effect is real, in spite of the efforts of the skeptics. I am impatient to see the new fire revolution really take off. Thank you for your dedication to this breakthrough. I wish you every success.
Paul
Dear Pekka Janhunen:
Thank you for your insight.
Warm Regards,
A.R.
Dear Pekka Janhunen,
Maybe the phenomenon that you describe, the evaporation of Nickel metal, is used in the E-Cats to create on a continuous basis enough grains of the right shape and size of the powder. Not all the Nickel grains do react at the same time. Maybe size matters, and after enough vaporization, every grain passes the right condition.
Give it enough time, and we will all understand.
Kind regards,
Koen
Dear Andrea Rossi,
Some thoughts of mine (not necessarily relevant) about longevity of high-temperature reactor versions.
All metals vapourise to some extent at high temperature. This is quantified as vapour pressure, e.g. http://www.powerstream.com/vapor-pressure.htm. Also, the flux of gas (atoms per time per area) is given by Jeans formula n*sqrt(kB*T/(2*pi*m)) where n=P/(kB*T) is the gas number density (1/m^3), kB is Boltzmann’s constant 1.38e-23 J/K, T is temperature in Kelvin and m is the mass of the atom in question (for example nickel). Nickel at 1000 C evaporates at about 100 atomic layers per hour which is a large amount (http://www.dtic.mil/dtic/tr/fulltext/u2/768332.pdf). This is a worst case estimate, i.e. one assumes that in the same chamber with the hot nickel surface is some other surface which is clearly colder which cold-traps the vapourised metal atoms.
I am not sure which is a worse threat for the LENR phenomenon: slow evaporation of the nickel microparticle fuel, or slow condensation of other substances on those particles. Given that the nickel fuel is probably (at least most of the time) the hottest component of the reactor chamber, nickel would be expected to mainly migrate away from it and deposit on other, cooler surfaces of the chamber. This process could be minimised by having different walls of the reactor at nearly the same temperature (which should in principle be possible to achieve by building the reactor somewhat larger and with less fuel so that less cooling power per area is needed).
On the other hand, if erosion of the nickel is not a problem, but deposition of other elements on top of of it is, then having cooler surfaces facing the gas volume should actually be helpful, as would also be using nickel or other “good” element as wall materials also in other than fuel regions.
Perhaps one of the keys to long-lived reactor is to consider metal vapour behaviour carefully and maybe try and influence it by the selection of shape, temperature and materials of all walls which face the same volume as the nickel microparticle fuel. If one wants to avoid other elements than nickel from depositing on top of the nickel, then all walls should be made of nickel, or alternatively of tungsten which has low vapour pressure. Or perhaps use some kind of bipolar reactor where the hot region is moved, by design, in an alternating fashion from one wall to the other and back again, and nickel migration takes place mainly back and forth between the two surfaces (similarly to how CO2 evaporates and condenses from south pole to north pole and back again every year on Mars).
regards, pekka
Dear Giuseppe:
the delivery, I suppose.
Warm Regards,
A.R.
Dear gio:
Only the low T 1 MW plant. The Others are prototypes to be suudied and certified.
Warm Regards,
A.R.
Joe wrote in April 30th, 2013 at 12:52 AM
1. You gave the example of a positive magnetic moment inducing a positive magnetic moment. A negative one inducing a negative one. Therefore, would not a null magnetic moment (that of 2He4) induce a NULL magnetic moment if it could orbit the z axis?
RESPONSE:
Yes, the magnetic moment induced by the magnetic field of 2He4 is NULL, since its TOTAL magnetic moment is zero.
However you had asked to me if the CHARGE of the 2He4 is able to induce magnetic moment.
And the response is yes, in the case of the 2He4 to have an orbiting motion.
Look at the calculation of the magnetic moment of 3Li6 in the Fig. 18, page 23.
I had used a phenomenological method of calculation because I dont know the speed of deuteron orbiting the central 2He4. So I could not calculate the contribution of the charge of the deuteron moving about the 2He4.
The value ∆= +0,035 in the Fig. 18 is due to the rotation of the magnetic moment µ= +0,857 of the deuteron and its charge q=1,6×10^-19C also orbiting the 2He4.
Therefore, that value ∆= +0,035 is due to the charge and magnetic moment of the deuteron working together, orbiting the central 2He4 in the 3Li6.
However I had concluded that the contribution of the charge is despicable face to the contribution of the magnetic moment, because if you turn the deuterons (and neutrons) upside down within the light nuclei, the magnetic moment of the nucleus change its sign and keep the same value. However the contribution of the charge do not change, because the charge continues moving in the same direction, but its contribution do not change the final value of the magnetic moment of the nucleus.
If the contribution of the charge should be considerable, it would have to change the value of the magnetic moment of the nucleus when the deuterons are turned upside down.
2. If 2He4 can induce a magnetic moment, then it should induce a magnetic moment in the 5B10 nucleus since the 2He4 would now be orbiting a barycentre that is common to itself and the three deuterons.
RESPONSE:
Yes, because of the rotation of its CHARGE.
However the contribution of the charge is despicable face to the magnetic moment of the 3 deuterons.
Regards
wlad
Did you start loading on the truck? Don’t forget to take a few pictures.
Post di Luca Salvarani.
Perchè! Domani cosa succede!!!
Dear Dr Rossi
i’m sorry if i’m becoming a little bit boring;
the Usa Partner will use the first three plants or one fo these ,for his heat centralized distribution facility.
Is this beginnig, a part of his sustainability mission ?
Warm regards
gio