# F&P's experiments – 30 years after CF announcement

• Do you agree the mathematical expression for molar rate of D2O vapor in a electrolytic cell is correct or not?

Mv / (Mv+Md+Mo) = P / P' .................................... (1)

Please, I just proved mathematically that this is perfectly correct.

Do you really not understand the math?

I tried to make the math as clear as possible. Is there anything more I need to explain in my details?

Read my comment once more and ask any question you would like

I had already told you (1) that "In your math, you repeated what I had already explained in my last jpeg, so no problem believing you, but this only applies to the vapor that saturates the gas bubbles produced by electrolysis."

You see? I too try to write as clear as possible. But, just in case, I try to explain what it means with other words.

It means that I believe in all your equations from [1] to [4] (*), but only if they are correctly applied for calculating the sensible and latent heat carried away by the vapor that saturates the electrolytic gas, which doesn't include the vapor produced by direct boiling.

(*) I use the square brackets to distinguish equations from links

Ascoli, please remember that the energy input in an electrolytic cell IS electricity.

Electricity provides the energy both to heat the water AND to produce electrolytic gas.

And electrolyte is the resistor where some of the electric energy is lost as heat.

Are you of the opinion that there are other energy sources than electricity that produced the bubbles ?

But of course is we have some excess heat produced by LENR then...

No, I'm of the opinion that, near or at boiling, the heat from electricity, which is the only form of energy required to explain the results reported in the 1992 paper (and IMO in all the other LENR reports), is mainly removed by the latent heat associated to the steam generated by two completely different mechanisms: (a) the saturation of the gas bubbles produced by electrolysis and (b) the vapor bubbles produced by direct boiling on the hot surfaces of the electrodes, the latter being by far the most important. The "Enthalpy content of the gas stream" term in the calorimeter model described in equation [1] of (2), as well as in Appendix 3 of (3), accounts only for the enthalpy carried away by the vapor produced by the first mechanism.

If you replaced the cathode with a resistor and the Anode with a resistor, you would still run current through the electrolyte, and (4) would work.

If you replace the anode and cathode with single resistive heating, there would be no current going through the electrolyte and (4) would be zero BECAUSE (4) is developed for an electrolytic cell NOT for pure resistive heating.

I was talking about the second case, of course, and you confirm that your equation [4] can't accounts for the vapor produced by Joule heating. You should remember that at boiling conditions, most of the electric energy is dissipated by Joule effect in the electrolyte and in the electrodes.

Please note that the formula for total molar rate of D2O vapor rate -

Mv= (3/4) * (I/F) * P / (P'-P) ..................................... (3)

Gives very differenet results at 40 degC and at 80 degC. Why? Because the vapor pressure P increase from 10 kPa to 50 kPa over the temperature range.

I know, but this variation has nothing to do with boiling. Consider these two opposite situations:

A) when cell temperature increases from 40°C to 60°C the vapor content in gas bubbles triples (the saturation pressure goes from 78.5 mbar to 245 mbar), but in both cases there is no boiling at all;

B) at boiling condition, if the cell voltage (V) increases by 10%, the current (I) also increases by 10%, it means that the gas stream increases by 10% and hence the vapor carried away by them increases by 10%. However, the electric power (V*I) dissipated inside the cell increases by 21% and, since the water is already at his maximum temperature, all this extra power should be removed by increasing the total vapor flux by 21%.

Do you see why the vapor produced by these two mechanisms (the saturation of the electrolytic bubbles and the direct boiling) are completely decoupled?

• You see? I too try to write as clear as possible. But, just in case, I try to explain what it means with other words.

Ascoli65 has no MEASUREMENT, NO EQUATION and NO CALCULATION .

Only PIXELS and VERBIAGE

In contrast to Fleischmann and PONS

who used MEASUREMENT, EQUATION, CALCULATION

to show excess heat from Pd/D2 electrolysis 30 years ago,

far in excess of chemical enthalpy change

• So, if you are confirming that the F&P paper on their 1992 boil-off experiment is wrong, we can take into consideration another one. But, I'm engaged with oystla to first discuss with him the 1990 seminal paper.

Ascoli! We all fully understand that for a unsuccessful hot fusion adept like you, that spent his live for the nonsensical idea to fuse Hydrogen by adding infinite amounts of energy idea, the P&F paper is shocking.

If your fear - and this fear is correctly based on the easily to predict sad future of ITER - is a consequence of your uncertain future, then please look for medical advise.

Even if you make some income out of your posting, it's not worth spending your live on refuting the only physical path which allows efficient fusion of Hydrogen.

Because you obviously miss some deeper understanding of nuclear physics - like most STDM adepts - you will have no chance to understand why LENR works. Instead of desperately trying to help a dying (hot fusion) community to survive one more year, you should realize, that if you are an early adapter, trying a new path, could help your situation much more than inventing sticky water bubbles...

• Ascoli! We all fully understand that for a unsuccessful hot fusion adept like you, that spent his live for the nonsensical idea to fuse Hydrogen by adding infinite amounts of energy idea, the P&F paper is shocking.

Hot foam and pixel adept more likely

Don't CALCULATION and EQUATION

exist in HOT FUSION engineering?

Ascoli65 exhibits no sign of

either CALCULATION OR EQUATION..

• consequences on the calculation of the energy balance

At least MIT and the 1989 hot fusion club had

MEASUREMENT, CALCULATION and EQUATION in 1989.

Ascoli only has PIXELS and VERBIAGE in 2019

Why?

MEASUREMENT and CALCULATION such

as when MIT manipulated the MEASUREMENT data

to show zero excess heat with Pd/D2O

can be easily refuted

That is WHY ASCOLI shows NO EQUATION,

NO MEASUREMENT

NO CALCULATION

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

• Even if I shouldn't reply to an OT and ad-hom comment like yours, I remind you once again that I've already explained my position on this subject (1).

As for the F&P papers, I think you could provide a better help to their scientific reliability if you explain how their calorimetric model takes into account the enthalpy losses due to the vapor produced inside their cell by direct boiling. Not a hard task for someone who is going to revolutionize the basis of the nuclear physics.

• As for the F&P papers, I think you could provide a better help to their scientific reliability if you explain how their calorimetric model takes into account the enthalpy losses due to the vapor produced inside their cell by direct boiling. Not a hard task for someone who is going to revolutionize the basis of the nuclear physics.

As a hot fusion adept as you are it is difficult to fully understand electro chemistry. The calorimetry P&F did was never intended to cover the boil off phase and you take this as a pretext to fulfill your blind hopes that P&F were wrong.

The heat estimate of the boil-off was always an extrapolation from the stable model of steady state electrolysis to the final total evaporation.

Instead off trolling us with with your nonsensical questions, you should try to educate your hot-fusion team members about the real back ground of nuclear physics. Especially try to elaborate with them that a simple hot plasma is the most unlikely state to get a working fusion reactor. There are more clever ways to do hot fusion like Mills & Lipinski do.

• As a hot fusion adept as you are it is difficult to fully understand electro chemistry. The calorimetry P&F did was never intended to cover the boil off phase and you take this as a pretext to fulfill your blind hopes that P&F were wrong.

The heat estimate of the boil-off was always an extrapolation from the stable model of steady state electrolysis to the final total evaporation.

Instead off trolling us with with your nonsensical questions, you should try to educate your hot-fusion team members about the real back ground of nuclear physics. Especially try to elaborate with them that a simple hot plasma is the most unlikely state to get a working fusion reactor. There are more clever ways to do hot fusion like Mills & Lipinski do.

Let there be no doubt

The calorimetry model F&P developed was perfectly valid at all temperatures.

Their black box model is a valid model for 40 degC as well as 100 degC.

This was also discussed in the Wilson Critique [2] and Fleischmanns answer [3]

And I proved that ALL D2O was covered in the model in this respons

F&P's experiments – 30 years after CF announcement

• It means that I believe in all your equations from [1] to [4] (*), but only if they are correctly applied

Dear Ascoli,

You can not both agree on formula (1) AND maintain your view that not all D2O vapor was included in Fleischmanns calorimeter model.

Let me explain in further details (and ask me any questions you would like) :

Again, the expression for TOTAL massrate of saturated D2O vapor leaving the electrolytic cell is

Mv / (Mv+Md+Mo) = P / P' .................................... (1)

where

Mv = Molar mass rate of Heavy water vapor leaving the cell

Md= Molar mass rate of D2 gas leaving the cell

Mo= Molar mass rate of O2 gas leaving the cell

P= Vapor Pressure

P'=Atmospheric pressure

The formula (1) is valid for all temperatures.

SO if you agree that (1) is the equilibrium formula for saturated D2O vapor then -

We may now rearrange (1) and solve for Mv

Mv= (Md+Mo) / (P'/P - 1) ...........................................(2)

And (2) will give you the TOTAL amount of saturated D2O vapor if you are able to

- measure Atmospheric pressure

- vapor pressure can be fund with knowledge of the liquid temperature

- Find the amount of D2 and O2 molecules leaving the cell

Do you agree?

F&P's experiments – 30 years after CF announcement

• I was talking about the second case, of course, and you confirm that your equation [4] can't accounts for the vapor produced by Joule heating. You should remember that at boiling conditions, most of the electric energy is dissipated by Joule effect in the electrolyte and in the electrodes.

Actually, if you read a little about water electrolysis you will learn that the Efficiency of splitting water increase with increasing temperatures. That is why there are a lot of work at developing high temperature electrolyzers.

i.e. There are relatively spoken more H2/D2 gas produced at 100 degC than at 40 degC

All electrolyzers have losses, i.e. the energy is both gone into D2 / O2 production AND heating the electrolyte.

For an electrolytic cell the formula (4) will produce the total amount of D2O vapor.

TOTAL Saturated D2O vapor energy rate stream = L * (3) = L * (3/4) * (I/F) * P / (P'-P)........................................... (4)

Ref my previous message above for the important formula (1) that is the starting point here.

The formula for D2O steam at pure resisitive heating of water would be a little simpler, but here we developed the formula for an electrolytic cell.

F&P's experiments – 30 years after CF announcement

• at boiling condition, if the cell voltage (V) increases by 10%, the current (I) also increases by 10%, it means that the gas stream increases by 10% and hence the vapor carried away by them increases by 10%. However, the electric power (V*I) dissipated inside the cell increases by 21% and, since the water is already at his maximum temperature, all this extra power should be removed by increasing the total vapor flux by 21%.

OK, I see another misunderstanding.

You think there is no additional D2 / O2 production even if current increase, but that's wrong.

Also D2+O2 rate increases as per

Md+Mo= I / (2*F) ..................................................... (2)

You did say you agreed with the formulas

Ref.

F&P's experiments – 30 years after CF announcement

• , which doesn't include the vapor produced by direct boiling.

Ascoli65 writes repetitively about the 5th Term.... the vapor term.

But Ascoli65 will not write down what the fifth term vapor is.?

What could vapor mean? D20 ?

Maybe Ascoli65 believes that there is secret D2O not included in Term 3 and Term4.

This D20 is too important and confidential to make it explicit in Term5

But perhaps it is A2O- Ascolian vapour? or A65O?

• I agree if your

"Mv = Molar mass rate of Heavy water vapor leaving the cell"

refers only to the vapor which saturates the streams of D2 and O2 bubbles produced by electrolysis, in accordance with Appendix 3 of the F&P's 1990 paper (1), where it is specified that: "the gas stream has been assumed to be saturated with D2O vapor at the partial pressure P which applies to the cell temperature". It means that the [A3.2] term of (1) accounts only for the steam that leaves the cell inside the D2 and O2 bubbles, and only within the limit of the quantity which saturates those bubbles. It does NOT represent ALL the vapor leaving the cell, in particular the vapor produced by direct boiling, which onsets when the water temperature is approaching the boiling point.

Actually your equation [2] expresses an upper limit, ie the "=" should have been a "=<". The reason is that, a low temperatures, the evaporation from the surrounding water is limited by the kinetic, ie by the evaporation rate. In the opposite situation, when direct boiling begins, your equation [2] makes no sense, because the denominator (P'/P-1) goes to zero.

Actually, if you read a little about water electrolysis you will learn that the Efficiency of splitting water increase with increasing temperatures. That is why there are a lot of work at developing high temperature electrolyzers.

i.e. There are relatively spoken more H2/D2 gas produced at 100 degC than at 40 degC

Are you sure? FWIK, every second, electrolysis generates 0.5 moles of D2 and 0.25 moles of O2 for each Faraday (F) of current (I). The "gas stream" term in the F&P model has no dependency from the temperature. It has only a coefficient "gamma", equal or less than 1, defined as the "current efficiency of electrolysis".

Quote

All electrolyzers have losses, i.e. the energy is both gone into D2 / O2 production AND heating the electrolyte.

In the F&P cells, the Joule heating of the electrolyte is always predominant, especially at boiling conditions, when cell voltage is high.

Quote

For an electrolytic cell the formula (4) will produce the total amount of D2O vapor.

TOTAL Saturated D2O vapor energy rate stream = L * (3) = L * (3/4) * (I/F) * P / (P'-P)........................................... (4)

Ref my previous message above for the important formula (1) that is the starting point here.

NOT the TOTAL amount of D2O vapor, only the limited amount required to saturate the D2 and O2 bubbles. See my first answer in this post.

Quote

The formula for D2O steam at pure resisitive heating of water would be a little simpler, but here we developed the formula for an electrolytic cell.

Which would it be this simpler formula in your opinion?

OK, I see another misunderstanding.

You think there is no additional D2 / O2 production even if current increase, but that's wrong.

Also D2+O2 rate increases as per

Md+Mo= I / (2*F) ..................................................... (2)

You did say you agreed with the formulas

The misunderstanding is yours. Please read again my phrase. I wrote that the "gas stream increases by 10%". Of course, I referred to the electrolytic gas (D2 + O2) which increases by the same amount of the current. However, due to the simultaneous 10% increase of the voltage, the dissipated power increases by 21% and, hence, the TOTAL production of vapor increases by the same amount, ie more than twice the amount predicted by the "gas stream" term included in the F&P model.

• The misunderstanding is yours

Ascoli65 writes repetitively about the 5th Term.... the vapor term.

But Ascoli65 will not write down what the fifth term vapor is.?

What could vapor mean? D20 ?

Maybe Ascoli65 believes that there is secret D2O not included in TERM3 and TERM4.

This D20 is too important and confidential to make it explicit in TERM5

Maybe Ascoli65 will make the BIG REVEAL at MIT on March 23.

TERM 5. Is it D2O or A2O? Ascolian vapour?

Time in Italy 11.45 pm Time in MIT 6.45 pm

• Ascoli was online for about 30minutes but has gone away

perhaps he trying to find his Ascolian vapour TERM5.....

TERM 5. Is it D2O or A2O? Ascolian vapour?

Time in Italy 12.15 am Time in MIT 7.15 pm

lenr-forum.com/attachment/7928/

• New

Time in Italy 12.17 am Time in MIT 7.17 pm

About six days to go till til the 30th anniversary of Fleischmann and Pons public announcement of their first findings

Just a reminder of why many of the first attempted replications failed

from Stan Szpak- electro-chemist

Thanks so much to Ruby Carat

Also in memoriam ...Stan Szpak(1920=2016)

https://www.infinite-energy.co…es/pdfs/SzpakMemorial.pdf

If I should live so long....

• New

About six days to go till til the 30th anniversary of Fleischmann and Pons public announcement of their first findings

Six days for avoiding the risk of the consequences predicted by oystla :

 So you started by the bold and earth shattering claim of "it doesn't account for the vapor carried out by the bubbles directly produced by boiling!", which would be devastating to all of F&P's work. May be you did not understand the ramification of your claim?

And, for my understanding, the ramification can hardly spare those who supported this work.

Btw, I'm still waiting his reply to my answers to his last remarks.

• New

I still like calling it galvanic corrosion. That way they know it takes time and not just a paint job.

• New

Six days for avoiding the risk of the consequences

I hope Ascoli65 had a nice St Patricks Day in MIT or Italy.

It was not so grand in my homeland, NZ.

Best wishes for March 23 in Italy or MIT .

Btw, I'm still waiting his reply to my answers to his last remarks.

I am waiting for Ascoli65's algebraic definition of " vapor bubbles produced by boiling"

in Term5.

Is it D2O or A2O?

Time in Italy 2.14 am .Time in MIT 9.14 pm.