Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[mixent]

In reply to  Jed Rothwell's message of Thu, 27 Jun 2019 19:58:27 -0400:
Hi Jed,
[snip]
>As shown in Table 1, he adjusted the gas during the 111 day run, to see the
>effects of pressure. He kept an inventory of gas to measure loading. But
>this could be kept constant in a future test. Plus you could probably
>measure the helium (if any) more easily than with most cold fusion reactors.

If he was measuring loading, then he knows how much gas was absorbed into the
Ni. If he could also measure what was left after the run then we would know how
much was consumed, and could consequently calculate the energy / D atom.

As Jones has pointed out, fusion or baryon destruction would yield MeV amounts /
D atom, whereas any form of "super chemistry" would yield on the order of 100's
to 1000's of eV / atom.

There is a pretty large gap between keV & MeV, so the result could well be
indicative, or at least alter the chances of any given theory being correct.

Could you at least suggest it to him?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Axil Axil]

One possible energy source not considered is extraction of energy from the
vacuum through Hawking radiation. Hawking radiation coming from a phonon
generating Bose condensate  has recently been experimentally demonstrated
that could be producing an event horizon in which Hawhing radiation is
forming. In the mesh reactor, the Bose condensate could be formed from
phonon polaritons.

"  Jeff Steinhauer  of the
Technion in Haifa, Israel, generates a sonic horizon by accelerating a
fluid of rubidium-87 atoms to a supersonic speed. In 2016, Steinhauer made
headlines by detecting
 the acoustic
analogue of Hawking radiation
.
Quantum units of sound, called phonons, popped up in pairs straddling the
sonic horizon; one phonon would get swept along by the moving fluid while
the other fought its way upstream and escaped. "

[image: Infographic titled "A BLACK HOLE ANALOGY", depicting
characteristics of a black hole such as quantum fluctuations, black hole
evaporation, hawking radiation. Below is a sonic black hole, depicting its
setup of atomic flow and sonic hawking radiation.]

See

https://www.quantamagazine.org/philosophers-debate-new-sonic-black-hole-discovery-20190625/


Philosophers Debate New ‘Sonic Black Hole’ Discovery

On Thu, Jun 27, 2019 at 8:23 PM JonesBeene  wrote:

>
>
> Robin,
>
>
>
> It would be almost impossible to be sure that the amount of gas was
> constant since the pressure varies during operation and small leaks are
> inevitable. Plus - where is gain coming from?
>
>
>
> Mass is being converted into energy but how? That method would be critical
> to know to calculate the energy per D atom. The main contenders look like
> this:
>
>1. D+D -> He
>2. Loss (deflation) of electron mass-energy-  Millsean
>3. Disintegration of deuterons into muons – Holmlid - which is more
>energetic than fusion
>4. Sequential oscillation of Coulomb explosions – emerging hypothesis
>from Hora, Miley etc.
>5. Any combination or permutation of the above
>
>
>
> If fusion of D into He is your choice - then one gram of deuterium yields
> 10^12 J (terajoules) of energy, so if there is a milligram in the reactor,
> it could generate about 278 kWhr but the internal pressure would need to be
> maintained, so that makes it difficult to quantify rate of consumption.
>
>
>
> If Mizuno was using close to 3 kW to heat his house, he could run it for
> only about 100 hours without a refill but  to maintain the internal
> pressure it would need constant pressure readjustment
>
>
>
> I do not believe that meaningful information can be obtained without
> mass-spectrometry of the contents after a long run.
>
>
>
> After 100 hours there should be a whopping milligram of helium in there –
> that should be quite  easy to measure in contrast to the puny subwatt
> reactors which have been responsible for the belief that nuclear fusion
> into helium is the main gain.
>
>
>
> My bet is that Mizuno will find almost no helium.
>
>
>
>
>
> *From: *mix...@bigpond.com
>
>
>
> If the amount of gas in the device is constant, then it should be possible
> to
>
> calculate the amount of energy generated by each D atom. That could point
> the
>
> way to an explanation. Has this been done?
>
>
>
> Regards,
>
>
>
>
>
> Robin van Spaandonk
>
>
>
> local asymmetry = temporary success
>
>
>
>
>

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Jed Rothwell]

JonesBeene  wrote:


> It would be almost impossible to be sure that the amount of gas was
> constant since the pressure varies during operation and small leaks are
> inevitable.
>

Nope. There are no measurable leaks. When there is no reactant in the cell
and it is pressurized at some low level, around 1000 Pa, you can leave it
for weeks and it does not gain more than a few Pascals.

When there is reactant, the pressure may decline by 10 or more Pascals per
day as it absorbs gas. Sometimes pressure increases as gas escapes from the
reactant.



> If fusion of D into He is your choice - then one gram of deuterium yields
> 10^12 J (terajoules) of energy, so if there is a milligram in the reactor,
> it could generate about 278 kWhr but the internal pressure would need to be
> maintained, so that makes it difficult to quantify rate of consumption.
>

The only way to detect this would be with a mass spectrometer.

- Jed

RE: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[JonesBeene]

Robin,

It would be almost impossible to be sure that the amount of gas was constant 
since the pressure varies during operation and small leaks are inevitable. Plus 
- where is gain coming from?

Mass is being converted into energy but how? That method would be critical to 
know to calculate the energy per D atom. The main contenders look like this:
1) D+D -> He
2) Loss (deflation) of electron mass-energy-  Millsean
3) Disintegration of deuterons into muons – Holmlid - which is more energetic 
than fusion
4) Sequential oscillation of Coulomb explosions – emerging hypothesis from 
Hora, Miley etc.
5) Any combination or permutation of the above

If fusion of D into He is your choice - then one gram of deuterium yields 10^12 
J (terajoules) of energy, so if there is a milligram in the reactor, it could 
generate about 278 kWhr but the internal pressure would need to be maintained, 
so that makes it difficult to quantify rate of consumption.

If Mizuno was using close to 3 kW to heat his house, he could run it for only 
about 100 hours without a refill but  to maintain the internal pressure it 
would need constant pressure readjustment

I do not believe that meaningful information can be obtained without 
mass-spectrometry of the contents after a long run.

After 100 hours there should be a whopping milligram of helium in there – that 
should be quite  easy to measure in contrast to the puny subwatt reactors which 
have been responsible for the belief that nuclear fusion into helium is the 
main gain.

My bet is that Mizuno will find almost no helium.


From: mix...@bigpond.com

If the amount of gas in the device is constant, then it should be possible to
calculate the amount of energy generated by each D atom. That could point the
way to an explanation. Has this been done?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Jed Rothwell]

 wrote:


> If the amount of gas in the device is constant . . .


As shown in Table 1, he adjusted the gas during the 111 day run, to see the
effects of pressure. He kept an inventory of gas to measure loading. But
this could be kept constant in a future test. Plus you could probably
measure the helium (if any) more easily than with most cold fusion reactors.



> , then it should be possible to
> calculate the amount of energy generated by each D atom. That could point
> the
> way to an explanation. Has this been done?
>

Not by me.

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Jed Rothwell]

Arnaud Kodeck  wrote:

The design of the reactor is huge to compare with the amount of heat.
>

Oh come now. Obviously it has not been optimized. It is 60 cm long and the
meshes are only 30 cm long. It is half empty.

It also has not been driven to maximum temperature. Mizuno has projected
that it could be far hotter with much higher power density.

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Axil Axil]

I have always thought that the gas insulator must be isotopically pure.
Deuterium is a lot easier to get off COTS compared to isotopically pure
prodium. This reflects the need to form a Bose condensate on the surface of
the mesh. This requirement for the formation of a condensate also reflects
why impurities such as water and nitrogen are a reaction show stopper. My
guess is that pure protium will work just as well as deuterium in the
Mizuno mesh reactor.

On Thu, Jun 27, 2019 at 4:18 PM JonesBeene  wrote:

>
>
>
>
> *From: *Jed Rothwell 
>
>
>
>- Two or three people have suggested to me that Mizuno's reaction must
>be unstable because it is exponential and self-heating. I do not think so.
>
> But Mizuno and other have suffered runaway reactions in the past which are
> completely unexplained to this day since they did not leave behind
> significant activation of materials..
>
> The best rationale for the belief that this design will not runaway is the
> very low inventory of reactant and/or the fact that the gainful reaction is
> not nuclear fusion.
>
> Not sure if the inventor or anyone else has calculated the inventory when
> it is operating  but it could be as low as a milligram, no?
>
> 300 Pa is about .003 atm. D2 gas weighs in at 4 g/mol. By Avogadro's law
> 1 mole of every gas occupies 22.4 L. For convenience we can guesstimate the
> volume of the reactor is 2.24 L  so that it holds one tenth of a mole.
> Therefore, the mass of D2 could be as low as .0012 grams during operation.
>
> OTOH -  a milligram of deuterium would definitely create a runaway if it
> all fused in a short time…
>
> ERGO – one big reason for thinking a runaway will not happen is that
> nuclear fusion is not the gainful reaction..
>
>
>
>
>
>
>
>
>

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[mixent]

In reply to  Jed Rothwell's message of Thu, 27 Jun 2019 17:08:27 -0400:
Hi,
[snip]
>> The best rationale for the belief that this design will not runaway is the
>> very low inventory of reactant and/or the fact that the gainful reaction is
>> not nuclear fusion.
>>
>Yes.

If the amount of gas in the device is constant, then it should be possible to
calculate the amount of energy generated by each D atom. That could point the
way to an explanation. Has this been done?

Regards,


Robin van Spaandonk

local asymmetry = temporary success

RE: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Arnaud Kodeck]

The design of the reactor is huge to compare with the amount of heat. For 
comparison, we had a SS cylindrical reactor (heated by inside) of 25cm length 
and 10cm diameter. It needs roughly 1kW to  keep it at 1000°C. @1000°C it needs 
more than 2W to increase the temperature of 1K. The question is if there is 
enough LENR reactions that the exponential overcome what the reactor can emit 
as radiation(T^4) and convection (becoming ‘negligible’ at first calculation 
for this kind of temperature in air).

 

Moreover there shall be an excitation inside the R20 (yet to be discovered) 
otherwise there will be no improvement with R19 and before. 

 

From: Jed Rothwell  
Sent: Thursday, 27 June 2019 20:52
To: Vortex 
Subject: [Vo]:Exponential and self-heating reactions are not necessarily 
unstable

 

Two or three people have suggested to me that Mizuno's reaction must be 
unstable because it is exponential and self-heating. I do not think so.

I commented on this at LENR-forum. Let me copy a few paragraphs.



First, you can see this is anomalous heat. Look at the first 10 minutes of Fig. 
6, and the stray points in Fig. 8. Resistance heating does not act that way.

Now look at the increase in Fig. 6 from minute 10 to hour 1:40. It is sedate. 
When resistance heating is turned up, the reactor heats about as quickly as a 
toaster oven. It takes a while for that increase to reach the flow calorimetry, 
but it does not take an hour and 30 minutes. This gradual increase is from 
anomalous heat. The anomalous heat increases exponentially in response to 
temperature, but "exponential" does not mean rapid, or out of control, or 
unstable. It just means the heat goes to a proportionally higher level in 
response to higher temperatures.

The reaction must be self-heating to some extent. 50 W of resistance heating 
alone would never begin to reach these temperatures. In Fig. 6, around 1:40 the 
heat leaving the reactor balances the heat being produced in the reactor, so 
the reaction stops going to higher levels. It stabilizes. Again, self-heating 
does not mean going out of control. Burning wood must self heat or the reaction 
stops, but that does not mean a wood fire goes out of control, or that it never 
reaches a terminal temperature and a stable, terminal heat production level.

 

 

 

. . . Apart from this, I have a feeling the heater is boosting the reaction for 
some reason other than just higher temperatures, such as IR stimulation. I will 
let others with more knowledge of physics speculate about that.

 

Anyway, I do not think an exponential self-heating reaction necessarily means 
the reactor might go out of control.

 

Re: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[Jed Rothwell]

JonesBeene  wrote:

But Mizuno and other have suffered runaway reactions in the past which are
> completely unexplained to this day since they did not leave behind
> significant activation of materials..
>
Right. But not with this system.

By the way, Biberian reanalyzed some of these reports. He thinks they may
have a conventional, chemical explanation.

Unexplained explosion paper:

https://www.lenr-canr.org/acrobat/BiberianJPunexplaine.pdf

Ah, ha. Maybe explained after all:

https://www.lenr-canr.org/acrobat/BiberianJPjcondensedy.pdf#page=81



> The best rationale for the belief that this design will not runaway is the
> very low inventory of reactant and/or the fact that the gainful reaction is
> not nuclear fusion.
>
Yes.

- Jed

RE: [Vo]:Exponential and self-heating reactions are not necessarily unstable

[JonesBeene]

From: Jed Rothwell

➢ Two or three people have suggested to me that Mizuno's reaction must be 
unstable because it is exponential and self-heating. I do not think so.
But Mizuno and other have suffered runaway reactions in the past which are 
completely unexplained to this day since they did not leave behind significant 
activation of materials..
The best rationale for the belief that this design will not runaway is the very 
low inventory of reactant and/or the fact that the gainful reaction is not 
nuclear fusion.
Not sure if the inventor or anyone else has calculated the inventory when it is 
operating  but it could be as low as a milligram, no?
300 Pa is about .003 atm. D2 gas weighs in at 4 g/mol. By Avogadro's law 1 mole 
of every gas occupies 22.4 L. For convenience we can guesstimate the volume of 
the reactor is 2.24 L  so that it holds one tenth of a mole. Therefore, the 
mass of D2 could be as low as .0012 grams during operation. 
OTOH -  a milligram of deuterium would definitely create a runaway if it all 
fused in a short time…
ERGO – one big reason for thinking a runaway will not happen is that nuclear 
fusion is not the gainful reaction..

Re: [Vo]: Mizuno's Q and A to a person who wants to replicate

[Jed Rothwell]

Mark Jurich  wrote:


> It is most likely that the R20 housing is simply a UHV Mass Spec Section
> and is an off-the-shelf component.
>

I do not think it is. Mizuno says he has a specialty supplier fabricate
these reactors according to his specifications.

[Vo]:Exponential and self-heating reactions are not necessarily unstable

[Jed Rothwell]

Two or three people have suggested to me that Mizuno's reaction must be
unstable because it is exponential and self-heating. I do not think so.

I commented on this at LENR-forum. Let me copy a few paragraphs.


First, you can see this is anomalous heat. Look at the first 10 minutes of
Fig. 6, and the stray points in Fig. 8. Resistance heating does not act
that way.

Now look at the increase in Fig. 6 from minute 10 to hour 1:40. It is
sedate. When resistance heating is turned up, the reactor heats about as
quickly as a toaster oven. It takes a while for that increase to reach the
flow calorimetry, but it does not take an hour and 30 minutes. This gradual
increase is from anomalous heat. The anomalous heat increases exponentially
in response to temperature, but "exponential" does not mean rapid, or out
of control, or unstable. It just means the heat goes to a proportionally
higher level in response to higher temperatures.

The reaction must be self-heating to some extent. 50 W of resistance
heating alone would never begin to reach these temperatures. In Fig. 6,
around 1:40 the heat leaving the reactor balances the heat being produced
in the reactor, so the reaction stops going to higher levels. It
stabilizes. Again, self-heating does not mean going out of control. Burning
wood must self heat or the reaction stops, but that does not mean a wood
fire goes out of control, or that it never reaches a terminal temperature
and a stable, terminal heat production level.



. . . Apart from this, I have a feeling the heater is boosting the reaction
for some reason other than just higher temperatures, such as IR
stimulation. I will let others with more knowledge of physics speculate
about that.

Anyway, I do not think an exponential self-heating reaction necessarily
means the reactor might go out of control.

[Vo]: Mizuno's Q and A to a person who wants to replicate

[Mark Jurich]

Arnaud wrote:
The reactor R20 is made of 2 standard conflat flanges and one pipe with 2 
conflats (on each side).  The flanges are available from the shelf of any 
vacuum supplier. On the other hand, the pipe is not not from the shelf. It 
needs to be built on demand. But maybe, this is from the shelf of the Mizuno’s 
supplier. In this case it would be nice to have the name of the supplier. So 
the replicator would have same SS tube (Size and composition).

Hi Arnaud:

FYI:

It is most likely that the R20 housing is simply a UHV Mass Spec Section and is 
an off-the-shelf component.  Many suppliers of Mass Specs will sell such a 
Section with a Mass Spec (I have an older Quadrupole Mass Spec that has almost 
the same dimensions for the housing) … If I find some additional time, I will 
respond back with a supplier or two, but nothing beats an exact response from 
Mizuno.

- Mark Jurich

RE: [Vo]:Fwd: Mizuno's Q and A to a person who wants to replicate

[Arnaud Kodeck]

Thank you for the answers.

 

I don’t agree about the heaters for 2 reasons. First, because if it is bent, EM 
generated is not equal to a straight inline heater. And secondly, the IR 
generated (and could be a stimulator of the reaction) depends on the type on 
heater it is. This cannot be ruled out so easily.

 

The reactor R20 is made of 2 standard conflat flanges and one pipe with 2 
conflats (on each side).  The flanges are available from the shelf of any 
vacuum supplier. On the other hand, the pipe is not not from the shelf. It 
needs to be built on demand. But maybe, this is from the shelf of the Mizuno’s 
supplier. In this case it would be nice to have the name of the supplier. So 
the replicator would have same SS tube (Size and composition).

 

Once the replication is successful, replicators can then change one variable at 
a time a confirm which variable is mandatory or not. In this perspective the 
idea to order the mesh (with Pd) directly is a good path for success. And also 
it can provide to Mizuno the necessary funding to repair other equipment broken 
by earthquake.

 

From: Jed Rothwell  
Sent: Tuesday, 25 June 2019 23:27
To: Vortex 
Subject: Re: [Vo]:Fwd: Mizuno's Q and A to a person who wants to replicate

 

Arnaud Kodeck mailto:arnaud.kod...@lakoco.be> > wrote:

 

Hi Jed,

 

Did you have by any chance some more information about the sheath heater? It is 
a 2m long, so how is the heater bent?  You said (I don’t remember if it is here 
or LENRForum) that you will ask the question to Mizuno.

 

I asked, but he has not had a chance to respond. When I asked him why he 
selected that heater, he said, "because it was the cheapest one."

 

It has to be 500 W and heat resistant at high temperatures. I expect any heater 
that meets those specifications will do. I doubt it matters. That is probably 
the first time I have said, "I doubt it matters" with regard to materials in a 
cold fusion experiment. Usually we have no idea what matters but we have to be 
fanatically careful about sourcing all materials.

 

 

Other thing, can Mizuno provide his supplier of SS316 tube ? The devil is in 
details.

 

I will ask. 


Mizuno has a local specialized company fabricate his reactors out of stainless 
steel. They are always bolted together not welded, as you see. They have thin 
metal gaskets. I do not recall what metal it is. The reactor has to be airtight 
enough to hold a vacuum for weeks, and it must not contaminate the gas at high 
temperatures. That is about all I know about it. I expect any stainless steel 
reactor that meets these specifications will do. I recommend you test it for a 
week, baking it out at low pressure and looking for contamination in the gas, 
using a mass spectrometer.

I do not know much about welding, but I do not think you should weld a reactor 
because I have heard the welds might leak, or contaminate the gas. I suggest 
you have a reactor fabricated by a specialized company. You might be able to 
find scientific supply houses that sell off-the-shelf reactors.  

 

I have a feeling this experiment is not fussy about the sources of the 
materials, or contamination. Mizuno is usually very careful about 
contamination, as are all electrochemists. He seems almost lackadaisical with 
this experiment. The method of making the reactant seems frightfully crude. I 
am amazed that it works at all. Yet it does seem to work. If it works, I'm 
happy, and I am not going to argue with Mother Nature about it.

- Jed