A simple electric heater, which has efficiency greater than 1

[George1]

Hi Leafy,
------------------------
Thank you for your reply.
Let us do the things in a step-by-step manner.
1) As a first step our theoretical water-splitting electrolysis OU concept must win public recognition in this forum.
2) And just after that we could talk about some revealing of our third secret, which allows drastically increasing of any standard battery's capacity.
Regards,
George

[George1]

To WaltzCee.
------------------------
Hi WaltzCee,
Thank you for your reply.
You wrote:"I think tax policy makers around the world have been using that formula or some variant of it for a very long time." We could not understand this. Would you be so polite to explain it in some more detailed manner?

[Leafy]

Hi George1,

I have an analogy for your theoretical electrolysis. Replace the electrolyte with an inductor. Apply energy to it. We get magnetic field energy + resistive heating of the coil. This is OU because magnetic energy is free in addition to omh’s law heating. Do you agree?

[Tarsier79]

And just after that we could talk about some revealing of our third secret, which allows drastically increasing of any standard battery's capacity.

Is your secret:

1. Another made up formula

2. Another misapplication of a mathematical formula

3. Another misinterpreted electrical law

4. Another incorrect assumption based on incorrect/incomplete data

5. Some other equally ignorant misconstruing of someones words

[George1]

1) We are talking solely and only about standard DC water-splitting electrolysis. And about nothing else.
---------------------------------------------
2) Anyway there is still no answer to our question: Is Prof. S. L. Srivastava's solution correct? Yes or no?
Only one word -- either "yes" or "no"!
KEEP WAITING FOR YOUR ANSWER. ONLY ONE WORD -- EITHER "YES" OR "NO".

[George1]

Let me remind again that there is still no answer to our question: Is Prof. S. L. Srivastava's solution correct? Yes or no?
Only one word -- either "yes" or "no"!
KEEP WAITING FOR YOUR ANSWER. ONLY ONE WORD -- EITHER "YES" OR "NO".

[George1]

There is still no answer to our question: Is Prof. S. L. Srivastava's solution correct? Yes or no?
Only one word -- either "yes" or "no"!

[MrTim]

Give it up, george1. You haven't convinced anyone here, except yourself (and many of us don't think you do either... ;-)

[WaltzCee]

spam

George's threads should moved to fraud.

[George1]

To WaltzCee and MrTim.
------------------------------------------
But why don't you answer my simple question? Why do you always avoid answering this simple question by using only either "yes" or "no"? Could you explain this to me and to all other members of this forum? (I strongly believe and hope that all we here in this forum are searching honestly for the truth.)

[Leafy]

I^2R is a theoretical calculation in your example. We don’t know if Ohm’s law apply in this situation. Have you ever see a resistor evaporate?

I blame it on the Asian country doesn’t promote enough crackpot forum for free thinkers to roam. But I must commend you mr.lt for going against the scientific norm. Your energy is extraordinary.

[George1]

Hi Leafy,
Thank you for your post.
I am answering your question immediately.
Please read carefully and thoroughly the text below.
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The text below can be found in many of our previous posts in this forum as well as in overunity.com. Anyway let us repeat it again.
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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=rrKFzL ... 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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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) 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 "&#1057;&#1073;&#1086;&#1088;&#1085;&#1080;&#1082; &#1079;&#1072;&#1076;&#1072;&#1095; &#1080; &#1074;&#1086;&#1087;&#1088;&#1086;&#1089;&#1086;&#1074; &#1087;&#1086; &#1092;&#1080;&#1079;&#1080;&#1082;&#1077;", 1986, p. 130, solved example problem 71. The authors of this book are &#1056;. &#1040;. &#1043;&#1083;&#1072;&#1076;&#1082;&#1086;&#1074;&#1072; and &#1053;. &#1048;. &#1050;&#1091;&#1090;&#1080;&#1083;&#1086;&#1074;&#1089;&#1082;&#1072;&#1103;. 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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[George1]

To Leafy.
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In one word, here is Prof. S. L. Srivastava's solution:
Current through the electrolyte is given by I = (m)/(Z x t).
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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It is absolutely evident from Prof. S. L. Srivastava's solution (a) that each second the electrolyte (the liquid conductor) consumes 31.86 J of electric energy and (b) that each second the electrolyte (the liquid conductor) generates 31.86 J of Joule's heat. Do you accept this simple obvious truth? Do you accept Prof. S. L. Srivastava's (and his two Russian colleagues' some 40 years ago) solution?
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Looking forward to your answer.

[Leafy]

I can agree that each second the electrolyte consumes 31.86J of energy. I do not agree that it generates 31.86J of heat each second. Now if you can prove that it generate 31.86J of heat(not through calculation and assumption) then I will believe you.

Matter of fact if you run DC through water and prove that the heating is equal the power input, then hydrogen must be free. Then I would accept.

[George1]

Hi Leafy,
You do not agree that it generates 31.86J of heat each second. Therefore you do not accept the validity of Prof. S. L. Srivastava's solution. Did I understand exactly your point of view?