Hi Floor,

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1) 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--------------------------

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?

YES OR NO?

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2) As for the balloon variation I perfectly agree with you -- it is really an OU device. Shall we do it? Some approximate calculations?

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Looking forward to your answers.

Regards,

George