The text below is a copy of our post of April 24, 12:50:36 PM. (The text below has been published many times in this forum topic.)

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Have a look again 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.

Prof. S. L. Srivastava's solution is given below.

Prof. S. L. Srivastava's solution consists of two lines only.

LINE 1. Current through the electrolyte is given by I = (m)/(Z x t).

LINE 2. Power consumed = (I) x (I) x (R) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) = 31.86 W.

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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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WE DEVELOPED FURTHER PROF. SRIVASTAVA'S SOLVED PROBLEM IN A NON-STANDARD MANNER.

OUR FURTHER DEVELOPMENT OF PROF. SRIVASTAVA'S SOLVED PROBLEM LED TO COP > 1.

HERE IS THE ESSENCE OF OUR APPROACH.

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1) Let us calculate the inlet energy, that is, inlet energy = (31.86 W) x (1200 s) = 38232 Ws = 38232 J.

2) The Joule's heat, generated in the process of electrolysis is given by

Q = (I) x (I) x (R) x (t) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) x (t) = (31.86 W) x (1200 s) = 38232 Ws = 38232 J = outlet energy 1.

3) 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 = (HHV) x (m) = (142 000 000) x (0.0001) = 14200 J = outlet energy 2,

where

m = mass of the released hydrogen

HHV = higher heating value oh hydrogen

4) Therefore we can write down the equalities:

4A) outlet energy 1 + outlet energy 2 = 38232 J + 14200 J = 52432 J

4B) inlet energy = 38232 J.

5) Therefore COP is given by

COP = 52432 J/38232 J = 1.37 <=> COP = 1.37 <=> COP > 1.

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IMPORTANT NOTE. 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 the same COP = 1.37, that is, we have again the same COP > 1.

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IMPORTANT NOTE. The text above must be evaluated (SOLELY AND ONLY!) by highly qualified experts (Ph.D.) in electric engineering. Otherwise nothing will come out of it.

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Looking forward to your comments.