A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[George1]

I am sending again our post of March 26, 2019, 10:39:21 AM.
"Please have a look at the book "Solved Problems in Physics", 2004, Volume 2, p. 876, solved problem 12.97. The author of this book is Prof. S. L. Srivastava (Ph.D.)
The same book can be found at the link  https://books.google.bg/books?id=rrKFzLB9KQ8C&pg=PA876&lpg=PA876&dq=%22electrochemical+equivalent+of+hydrogen%22&source=bl&ots=tQ8PSMLet3&sig=ACfU3U2HOLB78XHl2o3q-JanapzSK-McJA&hl=bg&sa=X&ved=2ahUKEwjDpp2-zZXhAhWT5OAKHUfuBzUQ6AEwBHoECAkQAQ#v=onepage&q=%22electrochemical%20equivalent%20of%20hydrogen%22&f=false
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For your convenience I am giving below the text of the problem and its solution.
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12.97. In the electrolysis of sulphuric acid solution, 100 mg of hydrogen is liberated in a period of 20 minutes. The resistance of the electrolyte is 0.5 Ohm. Calculate the power consumed. Electrochemical equivalent of hydrogen is 1.044 x 10 -8 kg/C.
Solution: The power consumed is equal to 31.86 W.
Prof. S. L. Srivastava stops here his calculations.
(The related solution's set of equations is not given here in order to save time and space. This set of equations however can be found in the book or in the link above.)
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The above solved problem has a potential which can be developed further. And here it is.
1) Let us calculate the inlet energy, that is, inlet energy = (31.86 W) x (1200 s) = 38232 Ws = 38232 J.
2) Let us calculate the current I. The current I is given by I = (m)/(Z x t) = 7.9 A,
where
m = 0.0001kg of hydrogen
Z = electrochemical equivalent of hydrogen
t = 1200 s
3) The Joule's heat, generated in the process of electrolysis is given by
Q = I x I x R x t = (7.9 A) x (7.9 A) x (0.5 Ohm) x (1200 s) = 37446 J = outlet energy 1.
4) HHV of hydrogen is 142 000 000 J/kg. Therefore the heat H, generated by burning/exploding of 0.0001 kg of hydrogen, is given by
H = (142 000 000) x (0.0001) = 14200 J = outlet energy 2.
5) Therefore we can write down the equalities:
5A) outlet energy 1 + outlet energy 2 = 37446 J + 14200 J = 51646 J
5B) inlet energy = 38232 J.
6) Therefore COP is given by
COP = 51646 J/38232 J = 1.35  <=>  COP = 1.35  <=>  COP > 1.
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Constant pure water and cooling agent supply could keep constant the electrolyte's temperature, heat exchange, mass and ohmic resistance, respectively.
Besides 0.0001 kg of hydrogen (and the related amount of the already split pure water) is small enough and can be neglected as a factor influencing the electrolyte's temperature, mass and ohmic resisitance.
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And one more interesting fact.
Literally the same solved problem can be found in an old Russian (still from the Soviet times) book "Сборник задач и вопросов по физике", 1986, p. 130, solved example problem 71. The authors of this book are Р. А. Гладкова and Н. И. Кутиловская. In the Russian version the data is a little different, that is, time is 25 minutes, the amount of generated hydrogen is 150 mg, Ohmic resisitance is 0.4 Ohm and the calculated power is 37 W.
Russians also stopped their calculations at 37 W.
Our further development of the Russian version led to COP = 1.37, that is, we have again COP > 1.
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Therefore the text above unambiguously shows that it is a matter of exact experimental data which is in perfect accordance with theory. Because I cannot imagine that three highly qualified experts in physics (yet strongly separated by time, space and nationality) would have made one and same mistake three times in a row. This is impossible!"
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Do you have any theoretical (ONLY THEORETICAL!) objections against the text above?
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Looking forward to your answers.
Regards,
George       

[George1]

Do you have any theoretical (ONLY THEORETICAL!) objections against the text of our previous post?

[Sergh]

1. If free energy is obtained with some types of electrolysis:
- it can be assumed that certain chemicals in the electrolyte can initiate the breaking of chemical bonds. There may be some aggressive elements that are in a chemically neutral compound. During electrolysis, neutrality is violated, and due to this, "zero point energy" is captured.  Maybe it's not sulfuric acid, maybe there was something else, as an impurity in the water from a conventional water supply system.

https://en.wikipedia.org/wiki/Redox

https://en.wikipedia.org/wiki/Electrophile

https://en.wikipedia.org/wiki/Nucleophile

2. If not.. At a current of 7.9 amperes and a voltage of 3.9 volts on the cell, strong electrolysis probably occurred. Probably the current could change over time. It was necessary to integrate the results over time. When using conventional U, I or Power meters, there may be a large measurement error occured.

http://www.designer-iii.com/cco/RMS.pdf

In Soviet times, obtaining efficiency> 100% was publicly ridiculed and was considered inexperience and experimental errors.
In principle, as elsewhere in the World. Remember what happened with Fleischmann - Pons:

https://en.wikipedia.org/wiki/Martin_Fleischmann

Therefore, no real scientist will write in an explicit form about obtaining an efficiency> 100%, because he will be afraid that his colleagues will laugh at him and consider him a marginal.

[sm0ky2]

(.....deep sigh)




I am not certain what your propaganda is trying to promote.
Are you selling books? Pimping out your favorite scientist?
What is the point of all of this?


Never mind


My point is simple.
How much Energy is required to bond a sulphur quadroxide to
the hydrogen molecule (H2) ?


Answer THAT, and we can talk about your "35% extra C.O.P."

[George1]

To Sergh.
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Hi Sergh,
Thank you for your reply.
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(Here is the beginning of your last post text.)
1. If free energy is obtained with some types of electrolysis:
- it can be assumed that certain chemicals in the electrolyte can initiate the breaking of chemical bonds. There may be some aggressive elements that are in a chemically neutral compound. During electrolysis, neutrality is violated, and due to this, "zero point energy" is captured.  Maybe it's not sulfuric acid, maybe there was something else, as an impurity in the water from a conventional water supply system.

https://en.wikipedia.org/wiki/Redox

https://en.wikipedia.org/wiki/Electrophile

https://en.wikipedia.org/wiki/Nucleophile

2. If not.. At a current of 7.9 amperes and a voltage of 3.9 volts on the cell, strong electrolysis probably occurred. Probably the current could change over time. It was necessary to integrate the results over time. When using conventional U, I or Power meters, there may be a large measurement error occured.

http://www.designer-iii.com/cco/RMS.pdf

In Soviet times, obtaining efficiency> 100% was publicly ridiculed and was considered inexperience and experimental errors.
In principle, as elsewhere in the World. Remember what happened with Fleischmann - Pons:

https://en.wikipedia.org/wiki/Martin_Fleischmann

Therefore, no real scientist will write in an explicit form about obtaining an efficiency> 100%, because he will be afraid that his colleagues will laugh at him and consider him a marginal.
(Here is the end of your last post text.)
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YOU ARE NOT READING CAREFULLY MY POSTS! PLEASE READ VERY CAREFULLY AGAIN WHAT I HAVE WRITTEN IN MY LAST POST!
The traditional scientists as shown in our last post are giving only a standard water electrolysis process problem. AND OUR TEAM DEVELOPED IT FURTHER AND DRAW THE CONCLUSION THAT STANDARD WATER ELECTROLYSIS PROCESS THEORETICALLY HAS EFFICIENCY/C.O.P. BIGGER THAN 1. That's all!
Looking forward to your answer.
George