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

Quote from oystla

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)

Quote from oystla

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

Quote from oystla

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.

Quote from oystla

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.

Quote from oystla

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?

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

(2) https://www.lenr-canr.org/acrobat/Fleischmancalorimetra.pdf

(3) https://www.lenr-canr.org/acrobat/Fleischmancalorimetr.pdf

Quote from Wyttenbach

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...

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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.

(1) Clearance Items

Quote from oystla

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?

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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.

(1) https://www.lenr-canr.org/acrobat/Fleischmancalorimetr.pdf

Quote from oystla

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.

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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?

Quote from oystla

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.

Quote from RobertBryant

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:

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.

Quote from sam12

https://e-catworld.com/2019/03…e-earth-man-bob-greenyer/

Your link points to a post on ECW titled "Primer for Kickstarter to Translate Alexander Parkhomov’s book – SPACE. EARTH. MAN. (Bob Greenyer)".

Can you please explain the connexion with the arguments we are discussing in this thread dedicated to the F&P experiments?