[Vo]:NY Times: The coal industry isn't coming back

[Jed Rothwell]

http://www.nytimes.com/2016/11/16/opinion/the-coal-industry-isnt-coming-back.html

This is an interesting analysis from M. Webber of U. Texas. There are some
interesting aspects of this which are not well known. For example, one of
the reasons East Coast coal companies are going out to businesses is
because years ago the railroad industry was deregulated. This lowered the
cost of shipping Western coal, which is "cleaner and cheaper then Eastern
coal." Webber points out that big energy companies such as Exxon Mobil are
opposed to coal. Quote:

Natural gas companies are the primary beneficiaries of, and now defenders
of, clean air and low carbon regulations. They include Exxon Mobil, the
world’s largest publicly traded international oil and gas company, which
operates in a lot of countries that care about reducing carbon emissions.
The company issued a public statement in support of the Paris climate
agreement on Nov. 4, the day it took effect. Shutting down coal in favor of
natural gas, which is cleaner and emits much less carbon, is a big business
opportunity for companies like Exxon Mobil.

In the battle between coal companies and major oil and gas producers, I
expect the latter will be victorious.


- Jed

[Vo]:just LENR INFO

[Peter Gluck]

http://egooutpeters.blogspot.ro/2016/11/nov-16-2016-lenr-info.html

peter

-- 
Dr. Peter Gluck
Cluj, Romania
http://egooutpeters.blogspot.com

[Vo]:How much helium?

[Jed Rothwell]

Years ago, Russ George told me that in one of his experiments he could "see
helium bubbles." At the time I said that is impossible because if you could
see the bubbles the reactor would be producing more power than any
laboratory experiment. However, yesterday I ran the numbers and found I may
be wrong about that. I have estimated that if the cell was producing ~1 kW,
perhaps he could see bubbles. This is not to say Russ was right. I do not
know how much anomalous power he measured. There could be bubbles from some
other source. I do not know whether he collected the gas and analyzed it.
But anyway, here is my estimate. I would appreciate it if readers here
would check the numbers.

Assumptions:

This is D + D fusion producing Helium-4.

The smallest bubbles you could see are fine ones. I suppose the total
volume of gas is approximately 1 cubic millimeter per minute. Anything
smaller would probably not be visible to the naked eye.

The gas in the bubbles is at STP. Probably not, but anyway, close to it.

D+D fusion produces 245,000 MJ per gram of deuterium. I double checked that
number with some web sites:

http://www.geo.cornell.edu/eas/energy/research_front_page/nuclear_fusion.html
http://electron6.phys.utk.edu/phys250/modules/module%205/nuclear_energy.htm
(U. Tennessee, value given in ergs per gram)
http://physics.ucsd.edu/do-the-math/2012/01/nuclear-fusion/

Also from U. Tennessee: Efficiency (E/mc2)  Chemical energy 3 x 10E-8%,
Fission 0.002%, Fusion 0.4%. In other words, the mass of helium is almost
the same as the starting mass of deuterium. I will ignore the lost mass.


Okay --

The volume of any gas at STP is 22.4 L per mole. One mole of helium weighs
4 g. So it weighs 0.1786 g/L.

1 mm^3 is 1/1,000,000 of a liter. 1 mm^3 per minute is 0.00016667 L/s.
So the helium weighs 0.2976 g.

Multiply that weight by 345,000 MJ gives 0.001026785714 MJ/s. That's 1,025
J/s (watts).

That's a lot less power than I thought!


Cross-check. Mel Miles says fusion produces helium at a rate of 10E11 to
10E12 atoms per second per watt. Divide Avagadro's number by 10E11 gives
1.66E-12 moles, or 6.64E-12 g helum, which multiplied by 345,000 MJ/g gives
2.3 W. Close enough!

Mel Miles measured at most ~400 mW of anomalous power as I recall. That
produces very little helium, as you see from these numbers. 0.400 J /
3.45E11 J = 1.16E-12 g/s (0.0012 nanograms). Right? He collected for 4,400
s, but still that's not many nanograms.


Interesting extrapolation. According to the International Energy Agency
(iea) "Key world energy statistics" the world primary energy supply in 2015
was 5,269 Mtoe. An Mtoe is "millions of tons oil equivalent." The
conversion factor is 41,868,000,000 MJ per Mtoe. So that's 2.21E14 MJ. It
would take 639 tons of deuterium to produce that with cold fusion. As I
said, I am ignoring lost mass, so that would produce ~639 tons of helium.

https://www.iea.org/publications/freepublications/publication/key-world-energy-statistics.html

In my book I estimated that you would end up with 1,227 tons of helium,
about twice as much. I do not know where the discrepancy came in. Perhaps I
calculated it wrong, or perhaps I was looking at the heat required to
produce the primary energy, with the assumption that all energy will come
from cold fusion powered heat engines. That issue is complicated because
not all of our energy today comes from heat engines. In the U.S. 15% comes
from hydroelectricity, wind and solar power, and 20% comes from nuclear
power. They do not usually include the raw heat from nuclear power in
estimates of primary energy.

The mass to energy conversion for 5,269 Mtoe is the same for any source of
energy, mechanical, chemical, fission or fusion. When you wind up a spring
driven clock it gains mass. As it runs down it loses mass. Some people have
the notion that only nuclear energy annihilates mass. That is incorrect.

Here is another extrapolation. A modern jet aircraft burns about 1 gallon
of kerosene per second. That produces 142 MW of heat which converts to ~90
MW of mechanical thrust with 63% efficiency. I think. So, assuming a cold
fusion engine is also 63% efficient, 4 g of deuterium would be consumed in
about 162 minutes (2.7 hours).

- Jed

Re: [Vo]:How much helium?

[Jed Rothwell]

Russ George  wrote:

Jed’s senility is showing in his recollection. In my work I have repeatedly
> shown helium bubbles, known as “loop punching” in the proper solid state
> science vernacular. These ‘bubbles’ form inside solid cold fusion metals.
>

Well, you did not specify. But anyway, that would be macroscopic. Visible.

If the power was high enough and the helium produced for a long time, it
would be macroscopic, as I estimated, no matter what sort of bubbles you
mean.

You are being disagreeable for no reason. Whyinsult me with comments about
senility when you did not specify the type of bubble, it was years ago, and
I just provided some evidence that you might be right?

- Jed

RE: [Vo]:How much helium?

[Russ George]

Jed’s senility is showing in his recollection. In my work I have repeatedly 
shown helium bubbles, known as “loop punching” in the proper solid state 
science vernacular. These ‘bubbles’ form inside solid cold fusion metals. They 
are perfectly consistent with ‘loop punching’ “bubbles” formed in nuclear 
active metals such as plutonium and californium where alphas are a product of 
abundant nuclear reactions. I confirmed my ‘loop punching’ “bubbles” in several 
of the finest labs in the world whose forte’ was the study of helium in metals 
formed under well-established nuclear processes, such as alpha production, 
helium ion implantation, and tritium decay.  My cold fusion work produced 
prodigious helium released from the metals as well as observed and confirmed by 
multiple top helium labs. Some of that helium released amounted to e16-e17 
atoms of it… do the math on how many joules/kilowatts that is. Jed’s long 
history of being an arrogant opinionated armchair curmudgeon, aka troll, has 
led him to disapprove of me and my work which began many years ago when we were 
in business together, I the inventor he one of the investors, and when the 
times got tough he was the first to begin back stabbing to try to protect his 
paltry investment when the time arrived in the business that is legendary to 
all venture capitalists, known as the “time to shoot the inventor”. If you 
believe anything Jed has to say I have a very nice used bridge near downtown 
NYC I can let you have cheap. 

 

From: Jed Rothwell [mailto:jedrothw...@gmail.com] 
Sent: Wednesday, November 16, 2016 2:35 PM
To: vortex-l@eskimo.com
Subject: [Vo]:How much helium?

 

Years ago, Russ George told me that in one of his experiments he could "see 
helium bubbles." At the time I said that is impossible because if you could see 
the bubbles the reactor would be producing more power than any laboratory 
experiment. However, yesterday I ran the numbers and found I may be wrong about 
that. I have estimated that if the cell was producing ~1 kW, perhaps he could 
see bubbles. This is not to say Russ was right. I do not know how much 
anomalous power he measured. There could be bubbles from some other source. I 
do not know whether he collected the gas and analyzed it. But anyway, here is 
my estimate. I would appreciate it if readers here would check the numbers.

 

Assumptions:

 

This is D + D fusion producing Helium-4.

 

The smallest bubbles you could see are fine ones. I suppose the total volume of 
gas is approximately 1 cubic millimeter per minute. Anything smaller would 
probably not be visible to the naked eye.

 

The gas in the bubbles is at STP. Probably not, but anyway, close to it.

 

D+D fusion produces 245,000 MJ per gram of deuterium. I double checked that 
number with some web sites:

 

http://www.geo.cornell.edu/eas/energy/research_front_page/nuclear_fusion.html

http://electron6.phys.utk.edu/phys250/modules/module%205/nuclear_energy.htm (U. 
Tennessee, value given in ergs per gram)

http://physics.ucsd.edu/do-the-math/2012/01/nuclear-fusion/

 

Also from U. Tennessee: Efficiency (E/mc2)  Chemical energy 3 x 10E-8%, Fission 
0.002%, Fusion 0.4%. In other words, the mass of helium is almost the same as 
the starting mass of deuterium. I will ignore the lost mass.

 

 

Okay --

 

The volume of any gas at STP is 22.4 L per mole. One mole of helium weighs 4 g. 
So it weighs 0.1786 g/L.

1 mm^3 is 1/1,000,000 of a liter. 1 mm^3 per minute is 0.00016667 L/s. So 
the helium weighs 0.2976 g.

 

Multiply that weight by 345,000 MJ gives 0.001026785714 MJ/s. That's 1,025 J/s 
(watts).

 

That's a lot less power than I thought!

 

 

Cross-check. Mel Miles says fusion produces helium at a rate of 10E11 to 10E12 
atoms per second per watt. Divide Avagadro's number by 10E11 gives 1.66E-12 
moles, or 6.64E-12 g helum, which multiplied by 345,000 MJ/g gives 2.3 W. Close 
enough!

 

Mel Miles measured at most ~400 mW of anomalous power as I recall. That 
produces very little helium, as you see from these numbers. 0.400 J / 3.45E11 J 
= 1.16E-12 g/s (0.0012 nanograms). Right? He collected for 4,400 s, but still 
that's not many nanograms.

 

 

Interesting extrapolation. According to the International Energy Agency (iea) 
"Key world energy statistics" the world primary energy supply in 2015 was 5,269 
Mtoe. An Mtoe is "millions of tons oil equivalent." The conversion factor is 
41,868,000,000 MJ per Mtoe. So that's 2.21E14 MJ. It would take 639 tons of 
deuterium to produce that with cold fusion. As I said, I am ignoring lost mass, 
so that would produce ~639 tons of helium.

 

https://www.iea.org/publications/freepublications/publication/key-world-energy-statistics.html

 

In my book I estimated that you would end up with 1,227 tons of helium, about 
twice as much. I do not know where the discrepancy came in. Perhaps I 
calculated it wrong, or perhaps I was looking at the heat required to produce 
the primary energy, 

[Vo]:How much helium?

[Jed Rothwell]

Russ George  wrote:

Jed’s senility is showing in his recollection. In my work I have repeatedly
> shown helium bubbles, known as “loop punching” in the proper solid state
> science vernacular. These ‘bubbles’ form inside solid cold fusion metals.
>

Did you confirm the gas in the bubbles was helium? If so, by what method?

- Jed

Re: [Vo]:How much helium?

[Jed Rothwell]

Brian Ahern  wrote:

> 1 mm3 is 1/100,000 of a leter not 1/1000,000 !
>
Nope, it is a million. There are 1000 cubic millimeters in a milliliter (10
x 10 x 10), and 1000 cubic milliliters in a liter. 1000 x 1000 = 1,000,000.

Confession: I am bad at arithmetic, so I also asked Mr. Google. Or try this:

http://www.convert-me.com/en/convert/volume/liter.html

- Jed

Re: [Vo]:How much helium?

[Jed Rothwell]

I wrote:


> Divide Avagadro's number by 10E11 gives 1.66E-12 moles, or 6.64E-12 g
> helum, which multiplied by 345,000 MJ/g gives 2.3 W. Close enough!
>

Oops. 1 mole of deuterium is 2 g. So that's 1.2 W. Even closer.

2 moles of deuterium fuse to form 1 mole of helium.

- Jed

Re: [Vo]:How much helium?

[Brian Ahern]

1 mm3 is 1/100,000 of a leter not 1/1000,000 !



From: Jed Rothwell 
Sent: Wednesday, November 16, 2016 5:34 PM
To: vortex-l@eskimo.com
Subject: [Vo]:How much helium?

Years ago, Russ George told me that in one of his experiments he could "see 
helium bubbles." At the time I said that is impossible because if you could see 
the bubbles the reactor would be producing more power than any laboratory 
experiment. However, yesterday I ran the numbers and found I may be wrong about 
that. I have estimated that if the cell was producing ~1 kW, perhaps he could 
see bubbles. This is not to say Russ was right. I do not know how much 
anomalous power he measured. There could be bubbles from some other source. I 
do not know whether he collected the gas and analyzed it. But anyway, here is 
my estimate. I would appreciate it if readers here would check the numbers.

Assumptions:

This is D + D fusion producing Helium-4.

The smallest bubbles you could see are fine ones. I suppose the total volume of 
gas is approximately 1 cubic millimeter per minute. Anything smaller would 
probably not be visible to the naked eye.

The gas in the bubbles is at STP. Probably not, but anyway, close to it.

D+D fusion produces 245,000 MJ per gram of deuterium. I double checked that 
number with some web sites:

http://www.geo.cornell.edu/eas/energy/research_front_page/nuclear_fusion.html
http://electron6.phys.utk.edu/phys250/modules/module%205/nuclear_energy.htm (U. 
Tennessee, value given in ergs per gram)
[http://electron6.phys.utk.edu/phys250/modules/module%205/images/nuclea3aa.gif]

Nuclear 
Energy
electron6.phys.utk.edu
Nuclear energy can be produced by either of two types of reactions, fission, 
the splitting apart of a massive atomic nucleus, and fusion of lighter nuclei 
into a ...



http://physics.ucsd.edu/do-the-math/2012/01/nuclear-fusion/

Also from U. Tennessee: Efficiency (E/mc2)  Chemical energy 3 x 10E-8%, Fission 
0.002%, Fusion 0.4%. In other words, the mass of helium is almost the same as 
the starting mass of deuterium. I will ignore the lost mass.


Okay --

The volume of any gas at STP is 22.4 L per mole. One mole of helium weighs 4 g. 
So it weighs 0.1786 g/L.

1 mm^3 is 1/1,000,000 of a liter. 1 mm^3 per minute is 0.00016667 L/s. So 
the helium weighs 0.2976 g.

Multiply that weight by 345,000 MJ gives 0.001026785714 MJ/s. That's 1,025 J/s 
(watts).

That's a lot less power than I thought!


Cross-check. Mel Miles says fusion produces helium at a rate of 10E11 to 10E12 
atoms per second per watt. Divide Avagadro's number by 10E11 gives 1.66E-12 
moles, or 6.64E-12 g helum, which multiplied by 345,000 MJ/g gives 2.3 W. Close 
enough!

Mel Miles measured at most ~400 mW of anomalous power as I recall. That 
produces very little helium, as you see from these numbers. 0.400 J / 3.45E11 J 
= 1.16E-12 g/s (0.0012 nanograms). Right? He collected for 4,400 s, but still 
that's not many nanograms.


Interesting extrapolation. According to the International Energy Agency (iea) 
"Key world energy statistics" the world primary energy supply in 2015 was 5,269 
Mtoe. An Mtoe is "millions of tons oil equivalent." The conversion factor is 
41,868,000,000 MJ per Mtoe. So that's 2.21E14 MJ. It would take 639 tons of 
deuterium to produce that with cold fusion. As I said, I am ignoring lost mass, 
so that would produce ~639 tons of helium.

https://www.iea.org/publications/freepublications/publication/key-world-energy-statistics.html
Publication: Key World Energy Statistics 2016 - 
iea.org
www.iea.org
Since 1997, the IEA has produced an annual compilation of its most used 
statistics in a booklet easily accessible to all – the aptly named Key World 
Energy ...




In my book I estimated that you would end up with 1,227 tons of helium, about 
twice as much. I do not know where the discrepancy came in. Perhaps I 
calculated it wrong, or perhaps I was looking at the heat required to produce 
the primary energy, with the assumption that all energy will come from cold 
fusion powered heat engines. That issue is complicated because not all of our 
energy today comes from heat engines. In the U.S. 15% comes from 
hydroelectricity, wind and solar power, and 20% comes from nuclear power. They 
do not usually include the raw heat from nuclear power in estimates of primary 
energy.

The mass to energy conversion for 5,269 Mtoe is the same for any source of 
energy, mechanical, chemical, fission or fusion. When you wind up a spring 
driven clock it gains mass. As it runs down it loses mass. Some people have the 
notion that only nuclear energy annihilates mass. That is incorrect.

Here is another extrapolation. A modern jet 

RE: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons

[Roarty, Francis X]

Bob , didn’t mean Casimir cavity per se but was trying to suggest the 
fractional hydrogen plasma loading deeper and deeper into the lattice powder 
inside the reactor expands into a larger area of Casimir like suppression that 
opposes the dilation direction of the muon. My rabbit hole was your initial 
post wrt to an office worker some distance from the reactor getting sunburned 
without explanation while techs and engineers working on the reactor remain 
unaffected. I was looking for some relativistic wormhole that might explain. In 
my initial investigation into similarity between skeletal cats of Mills and 
nano powders of Rossi I theorized the Casimir cavities and suppression geometry 
of Ni nano powders are inverses of the other and are equivalent but I prefer to 
take a neo Casimir perspective.  When a muon with SR delayed radioactive decay 
intersects my proposed Casimir like plasma it is suddenly inside an inertial 
frame that now accelerates the decay rate in opposition to the SR velocity of 
the muon. As always time doesn’t change from a local perspective but there is 
suddenly more distance available for the muon to continue forward inside the 
reactor from a local perspective while the plasma seems to keep shrinking away. 
I think we have an odd relativistic situation where SR dilation by virtue of 
the muons velocity slows time AND the vacuum suppression of the reactor 
accelerates time COMBINE to give the muon a strange temporal vector, if this 
was simple polar coordinate addition the opposing temporal additions would 
simply cancel and spatial location remains fixed but SR is a Pythagorean 
relationship between velocity thru space and time while suppression is only 
based on geometry of the surrounding environment the particle is passing thru. 
There is also a distinct difference in the type of Lorentzian contraction to be 
considered, SR has a single axis of contraction while suppression seems to be 
symmetrical. My point is that this might allow for your odd prediction of a 
safe spatial zone immediately surrounding the reactor and muons returning from 
a “temporal long way round” vector to poison the remote office worker?  Ok, 
after re-reading this is even a long shot for me but will still send so you 
don’t think I was suggesting the muon was traveling thru a few Casimir cavities 
–obviously we would have measured an anomalous decay rate a long time ago if it 
were that easy to deal with radioactive waste.


From: Bob Higgins [mailto:rj.bob.higg...@gmail.com]
Sent: Monday, November 14, 2016 8:09 PM
To: vortex-l@eskimo.com
Subject: Re: EXTERNAL: Re: [Vo]:Holmlid, Mills & muons

Hi Fran,
I am unable to imagine how something special would happen in that case.  A muon 
in slow motion may have a greater chance of interaction if its energy is near 
the ionization energy of the atoms upon which it is incident - but this is only 
a small energy - less than 10eV.  At higher energy, it is probably more likely 
that the muon is going to ionize the atom and then scatter at lower energy.  
The distances are so small in condensed matter that the scattering will happen 
rapidly and will reduce the muon to the sweet spot wherein it can interact with 
the chemical (electronic) structure of the next atom it meets.

How would a brief passage though a Casimir geometry alter these behaviors?

On Mon, Nov 14, 2016 at 2:12 PM, Roarty, Francis X 
> wrote:
Bob, what if the “muon” doesn’t have to achieve light speed but rather becomes 
so “suppressed” think traveling thru a tiny Casimir cavity that the muons 
actual speed inside the cavity where vacuum wavelengths are dilate by 
suppression appears to achieve negative  light speed relative to observers 
outside the cavity where vacuum wavelengths are not suppressed.. IMHO catlitic 
action is a weak cousin to Casimir action and the longer wavelengths we 
consider suppressed are actually still present from the perspective of a local 
observer in the cavity.. the calculations of decay and distance traveled are 
then complicated by their Pythagorean relationship to the spacetime inside 
these cavities traveling distances we instwead perceive as dilation… but not 
just the dilation from their spatial displacement, rather the cavities push 
this dilation in the opposite direction and to some extent cancel?
Always out on a limb,
Fran

RE: [Vo]:How much helium?

[Russ George]

Yes, of course why would anyone not do so. The methods used were all of the 
usual state of the art methods, just do your reading into the complexities of 
measuring helium in metals and you’ll see how it is done. It’s all at your 
Googling fingertips. 

 

From: Jed Rothwell [mailto:jedrothw...@gmail.com] 
Sent: Wednesday, November 16, 2016 3:45 PM
To: vortex-l@eskimo.com
Subject: [Vo]:How much helium?

 

Russ George  > wrote:

 

Jed’s senility is showing in his recollection. In my work I have repeatedly 
shown helium bubbles, known as “loop punching” in the proper solid state 
science vernacular. These ‘bubbles’ form inside solid cold fusion metals.

 

Did you confirm the gas in the bubbles was helium? If so, by what method?

 

- Jed

 

Re: [Vo]:How much helium?

[Jed Rothwell]

Okay, "cubic milliliters" is redundant. Like round circles.

[Vo]:Slate: How Dumping Iron in the Ocean Can Help Fight Climate Change

[Jed Rothwell]

An article featuring Russ George, but not in a good way:

http://www.slate.com/articles/technology/future_tense/2016/11/how_dumping_iron_in_the_ocean_can_help_fight_climate_change.html