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Name: Zorislav Casar
Feedback: You made a excellent work. Thanks for very useful information
from your site.
I've made aluminizing chamber for telescope mirror and planing to
modify it to fusor.
If you are interested there is some my photos on this sites:
It's in Croatian but pictures maters.
I hope you will help me in my task.
Thank you for your kind feedback!
Looking at your impressive photographs I’m sure that a fusor is a very
suitable new project for you.
Please let me know when I can be of any help to you.
Name: Zorislav Casar
Allow me to inform about my current fusor status, but first Sorry for
my bad english.
Since D2O is not cheap (I bought 2,5cc Deuterium Oxide 99,90% purity
from ebay for 9€)
I need something to produce minimum quantity of D2:
Making micro electrolyser capable to work with few drops of D20
(picture v0.1 below).
Built small PEM cell.
But I didn't find minimum quantity of D2O to soak PEM membrane for this
To reduce D2 consumption (since My chamber jar is pretty big - ID 30cm,
Height 50cm ) I'm thinking to replace it with smaller one.
Any suggestions for this my ideas.
You do not need to apologize for your English. It is very OK and I do
understand very well your questions.
With 2.5 cc D2O you can in theory produce about 2.5 liters of D2 gas.
The drawing of your micro electrolyses is very ingenious. The
disadvantage is however that you need to add baking soda (NaHCO3) to
your D2O to convert it into an electrolyte. This will produce H2 gas
impurity in your D2 gas, which is unwanted.
Please read my webpage about D2 electrolysis: http://www.fusor.eu/deuterium.html
The smallest PEM cell available can be found here: http://fuelcellstore.com/horizon-mini-pem-fuel-cell
with dimensions outside of 3.2 x 3.2 x 3.2 cm
The standard Horizon PEM cell needs about 5 cc of D2O. You may need two
bottles of 2.5 cc. Note: it is important that the PEM tissue always
remains soaked or the cell will be destroyed. Therefore after use store
the cell wet in a hermetically sealed plastic bag.
Fuel consumption by the Fusor in absolute volume per time unit is very little and rather independent of size: 1
SCCM (one standard cubic centimeter per minute). It is not the size of
the chamber that determines the amount of D2 consumed but the
absorption by the grid and the loss through the vacuum pump. The size
of the chamber is only permitting the length of the way that the
Deuterium ions travel and permit them to gain speed for fusion. You are
not saturating the total volume of the chamber with ions but
concentrating the ions with high speed due to high voltage potential
into the grid sphere area where they are supposed to fuse. When they
not fuse they will pass the grid (or hit the grid wire and be adsorbed
and go lost) and travel through the chamber for gradually loosing speed
and then reversing their direction and gaining speed again towards the
grid again and so on.
Note: A Deuterium gas flow of 1 SCCM introduces in the vacuum chamber
per minute a quantity of n=PV/RT = (1 atm)x(0.001 L)/(0.08206 L atm mol-1 K-1 x 273.15 K) = 4.46 x 10-5 mol. The gas constant R has a value of 8.314 J/mol K or 0.08206 L atm mol-1 K-1. Avogadro's number is used to convert from moles into molecules: 4.46 x 10-5 mol x 6.02 x1023 molecules/mol = 2.69 x 1019 molecules Deuterium per minute or (theoretically) 5.37 x 1019 Deuterium atoms per minute or 8.95 x 1017
Deuterium atoms per second, which may turn into ions when all Deuterium
atoms will become ionized. Ionization of Deuterium atoms depends on the
number of charges occurring in the vacuum chamber. A current of 1A is
equivalent to 1 Coulomb per second which is 6.24 x 1018 electrons per second. In our example fusor we apply a current of 10 mA, which equals 6.24 x 1016
electrons/sec, potentially creating an equal amount of ions per second.
Ions will be circulating a number of times, expressed as the
recirculation factor which may vary between 10 and 100. The total
amount of Deuterium ions present in the vacuum chamber at a given
timepoint will therefore be higher than the number of ions per second
formed by ionization. This calculation is, however, based on the flow
of Deuterium gas into the vacuum chamber. However, we know that the
vacuum in our example fusor is kept constant at a pressure of of 10-2
Torr (10 micron) and that our spherical fusor has a radius of 10 cm.
The volume of the vacuum chamber as a function of the radius is:
or 4.1888 x 103 = 4188.79 cm3 which equals 4.18879 L.
Since nV=P/RT and R = 62.3637 (L Torr)/(mol K), P = 0.01 Torr, T =
273.15 + 20 = 293.15 K and V = 4.18879 L we get n = (4.18879 L x 0.01
Torr)/(62.3637 L Torr K-1 mol-1 x 273.15 K) = 2.459 x 10-6 mols in the vacuum chamber, which equals 2.459 x 10-6 mol x 6.02 x 1023 (Avogadro's number) molecules/mol = 1.48 x 1018 molecules in the vacuum chamber or 2.96 x 1018 atoms Deuterium, which (may) get ionized.
We notice that with a flow of 1 SCCM of Deuterium gas, a pressure of 10
micron and a current of 10 mA, one or the other way we obtain roughly 1018 Deuterium ions per second in our Fusor.
With a constant flow of 1 SCCM Deuterium gas, roughly per second 1018
Deuterium ions will enter the vacuum chamber and randomly occupy the
volume of the vacuum chamber. The attracting forces of the cathode
compete more or less with the "suction force" applied by the vacuum
pump(s). This "suction force" is not to be taken litterally as the
pump(s) only constantly remove(s) gas particles which causes the
remaining particles inside the system to re-arrange themselves in order
ideally to get an equal distribution. At the same time more Deuterium
arrive into the system (over 1018 per second!) and an equilibrium
will be the result. This equilibrium can be observed from the fact that
the instream of Deuterium gas is regulated in such a way that the
pressure (vacuum) of the system is kept at a constant pressure.
Name: Zorislav Casar
after some thinking I'll try to build micro electrolyser but with
high voltage short pulse generator. This option will (drastically?)
reduce amount of additional salt.
(Google-d "Water Electrolysis with Inductive Voltage Pulses",...)
It will take some time to make this piece of hardware and after that
I'll inform about the result.
Thank you for informing me about the inductive pulse electrolysis
I was not yet aware of this technique, but i am an old school person!
Next weeks I shall update my website with this info.
Please keep me informed about your progress.
No update of the website has been made so far due to complexity of the
suggested method. Constructing a high voltage DC nano pulse generator
(HVNPG) requires a special static inductive thyristor (a SIThy), which
is manufactured by NGK in Japan but appears not to be available to
Concerning this HVNPG electrolysis method a search on the internet
yielded information with the following timeline:
Japanese team discovered/invented nano-pulse method, published article
in 2005-2006 (reference 1).
Japanese team receives 1st August 2006 United States Patent 7,084,528 B2
Latvian team analyzed pulse behavior in various electrolyte
concentrations, 2011 (reference 2).
Next, in 2012 an article was published in India with results (reference 3).
The researchers in india reported a power saving, compared to
conventional electrolysis methods, of 98.6%!
Most amazing, however, is the fact that with the HVNPG method the
so-called overunity world comes peeping around the corner. This fact
and the fact that the SIThy cannot be obtained makes it impossible (or
inconvenient) for me to publish it on this website.
For those who are interested to receive the concerned PDF files
(literature as mentioned in the timeline) that I have collected, please
ask me through the feedback submission modus on this page.
Name: Tom Clemmerson
Feedback: I'll be your first feedback :) I am the guy you got the small
PMT and LiI crystal from. The two things you need to get set up and
going first are the vacuum system and the HV system. Practice using
air, nitrogen, helium, then pure hydrogen for your fusor before using
your deuterium. Instead of the usual spherical geometry try a
tetrahederal wire frame geometry for the cathode. It will reduce
scattering and increase actual fusion. Oh btw, I have some green fast
neutron detection fibers currently.
Thank you for your feedback (and the PMT and LiI crystal 😄). I’m sorry
that it is the second feedback and not the first one, which was not yet
published due to the fact that I needed to sort out the high voltage
nanopuls electrolysis issue first.
Nevertheless your feedback is very welcome.
The vacuum system is almost complete, though a rotary vane pump is
still in the process of being overhauled.
The HV system gives me more problems. I do have the X-ray transformer
but it has a very low duty cycle. Using this transformer requires heavy
cooling of the transformer oil in order to make it operate for a
sufficiently long time. Therefore I shall need a cooling compressor
with a liquid/liquid heat exchanger. This is a general problem is
setting up the project: having realized one issue, another issue turns
up. BTW Any suggestions for a better and cheaper HV power system?
It is indeed my intention to practice with air, helium and hydrogen
prior to using deuterium.
I shall dive into your suggestion of a tetrahedral wire frame geometry
for the grid. This geometry is new for me. Is there any reference to
this geometry or is it your own experience?
I noticed the green fast neutron detection fibers on eBay and I’m
interested in assembling a fast neutron detection system. Is one fiber
sufficient for detecting neutrons? As you have a lot of stuff on eBay:
what do I need more (which PIN photo diode assembly?)
Name: Tom Clemmerson
The tetrahederal geometry effectively creates an potential well
gradient at the center of the device rather than an infinitesimally
small zeropoint focus with a zone of spherical aberration around it. As
you may be aware the main effect in these fusors is scattering with a
standard spherical geometry. Simply it is very hard to make a truly
perfect and uniform field with a non uniform wire cage. In a nutshell
using a tetrahedron wire frame will make the nuclei more likely to fuse
because the probability the nuclei hitting each other with the magnetic
field vectors lining up correctly is greatly increased. A couple
magnets from a magnetron aligned at the base and top of the tetrahedron
will increase the acceleration even more, but you'll have to find the
sweet spot as the increase also comes with deflection the stronger the
field gets. What you will end up with is a hybrid z-pinch/fransworth
To make the x-ray transformer run at a better duty cycle, you should
get a larger transformer say for a GE, Ritter or SS White x-ray head.
Using a dynarad field x ray transformer would be the best though since
it is designed to be actively cooled. The trick to cooling them is to
use a manifold with jets that actually make the oil flow past the
transformer. as for cooling the oil, a simple $30 transmission cooler
from an auto parts store and a couple of server fans will do the job
nicely. The one you have looks like a Orilix 70 dental machine
transformer, and yeah it is a bit difficult to cool. You will also need
to soak it in hot oil and vaccuum out the air and moisture or it will
self destruct- violently.
The neutron detecting fibers were used in an array in a device (64 of
them) that looked for atomic bomb material down to the low miligram
range from a distance of at least 6 meters. One or two fibers should do
fine for fusor neutron detection since it can yield a decent amount of
neutrons when it is reacting strongly. They would be best used in
combination with your LiI in a moderator as a confirmatory measure.
Your moderated LiI will detect first, then when the fiber starts
detecting, then you know you got fusion going in inside. They are best
read with an APD detector. The one i recommend for you is the unit with
the four wires, as they came from a device that used a laser to measure
the weight of very small particles and can sense a single photon when
cooled and well shielded. Another item I recommend is the CsI:Tl+
photodide low energy gamma detector to measure the x-ray output for
safety and other reasons like detection of sputtering arcing before any
real damage is done.
Anyway, I hope this information is helpful to you :)
Your information is very useful to me and it makes me think away from
the current path.
I have a feeling that the tetrahedron grid concept comes near to
Bussard’s patents for realizing standing waves of ions. Apparently it
should also be beneficial to adapt the outer vacuum chamber into a
similar tetrahedron or even octahedron shape instead of the traditional
spherical shape? Anyway, I shall need to dive deeply into this matter
as it is an interesting issue.
Your suggestion for the air cooled oil cooling device for the X-ray
transformer is also a good idea. The funny thing is that a transmission
cooler will be difficult to find in Europe. The majority of cars here
have manual gear boxes! Nevertheless, an oil or water cooler as used on
certain motorcycles can be found easily.
Concerning the fiber, the ADP and the CsI:Tl thin detector with
photodiode: I hope to be in business with you coming weekend!
Name: Roberto Ferrari
Congratulations for all your effort and great presentation.
I have been around forums and web pages-fusor related and I have a
What would be wrong using heavy water vapor as fuel?
Would the negative ions bother deuteron interactions?
Thank you for your feedback.
Those fusioneers who use heavy water to produce deuterium gas for their
Fusor always put a gas drier in the supply line to the Fusor to
eliminate heavy water vapor. Apparently, heavy water vapor in the Fusor
has a negative effect on Deuterium fusion.
Knowing this, it never occurred to me to try to find the scientific
rationale for this phenomenon.
Nevertheless it is known that the impact of electrons on heavy water
vapor will cause radiolysis of the heavy water and form a selection of
different ions: D2O+, D+, OD+, D3O+, O+, D2+, O2+, OD-, D- and O-. The
mentioned ions will not be present in equal quantities as this may
differ depending on the energy applied. The relatively huge amount of
different particles will obstruct fusion of D+ particles. In this
perspective we should note that for proper fusion of Deuterium gas it
is essential that the vacuum chamber of the Fusor is evacuated up to a
pressure (vacuum) of 10E-3 Torr (1 micron) and preferably better, i.e.
10-4 to 10-6 Torr (0.1 to 0.001 micron), followed by filling with
Deuterium gas up to a pressure of the vacuum chamber of 10-2 Torr (10
micron). This shows that a relatively high amount of D+ particles are
required to obtain fusion.
Note: see above the earlier answer where has been calculated that roughly 1018 ions Deuterium per second flow into the vacuum chamber.