A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[Floor]

Dear Floor,
You want to convince all of us here in this forum that if 6 = 1 + 2 + 3, then 6 = 1. Congratulations! Fantastic achievement! You are a founder of an entirely new and revolutionary branch of mathematics! Take your medicine!:) Triple dose!:)

@George1

If you see a math error, just point it out.  I don't see one, at least not on my part.
       and
no not    6 = 1 + 2 + 3, then 6 = 1
      but
6 / 3 = 2 and 6 / 3 x 3 = 6   very well then, if this will clarify it for you
(6 / 3) = 2 and (6 / 3) x 3 = 6

In most circumstances energy is "conserved".  This is what makes it seem (to some) that
any one looking for O.U. processes is nutty.

Looking for O.U. has been for me, a path of learning.  To my mind, seeking O.U. is not
the exploration of how to get something from nothing, but rather the exploration of how
we might get energy from new and / or possibly little understood or even unknown
sources or methods.

Have we missed something ?  Perhaps that the ordinary electrolysis of water is O.U. ?   

If we have missed it,  you for one, certainly have not demonstrated how.
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Analogous to water through a multi-path pipe system, electric current divides between
any 2 or more paths, in proportion to the resistance to its flow along those paths.

In the electrolysis process, the electric current's path is divided (within the electrolyte)
into paths where in some of the energy is expended as heat via conventional
electrical resistance and some of the energy is expended in splitting the water molecules.

Like unto the water and the water pipe analogy, the total current entering those paths, is
equal to the total current exiting them. 

In this case, the total energy in and out are equal as well.  But that energy has been
transformed into other than a purely electromagnetic form.

E or I alone, neither amounts to power.  But if there is a current flowing it is because an
electric potential has been realized. Where there exists an electric potential, current will flow
when the electrical resistance is low enough  (E / I x R).

Your, cut and past repetition of formulas (whether chemical or electrical) does not equate to
an understanding of those formulas.  If you have misunderstood them,  just get it straight
and move forward.  That's all.

If you make a few pennys from the views here, I don't begruge that.  But the forum is not
a venue for / of con men selling false free energy devices.

[George1]

We are talking solely and only about standard DC water-splitting electrolysis process.
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Whatever experiments to do you actually measure constant voltage V, constant direct current I, constant Ohmic resistance R and time t. The real experimental values of V, I, R and t would give three simple mathematical expressions:
1) V = I x R. This is Ohm's law.
2) V x I x t = I x I x R x t. This is the first Joule's law.
3) V x I x t < (I x I x R x t) + (Z x I x t x (HHV)). This is the OU property of any standard DC water-splitting electrolysis process.
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Simple, clear and understandable. And anyone, who tries to reject this simple obvious truth, is either a clumsy and unskillful manipulator and/or a person, who has some severe mental problems.

[Floor]

George1 doesn't know basic electrity very well.

So for anyone interested in learning the basic electricity required to understand
the energy transfer and transformation involved in the dc electrolysis of pure water...

In addition to ohm's law as

E = I x R  voltage = Amperage x Resistance
I x R = E
E / I = R
E / R = I

there is next also ......

Kirchhoff's first law

For any junction in an electric circuit, the sum of currents entering that junction is equal to the sum of currents exiting that junction.

Kirchhoff's second law

The sum of the electric potential (voltage) drops around any closed circuit is equal to zero.

https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws
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[Floor]

@ All readers

VERY NICE... complete course !

Basic electricty books. I have these in my own library (hard copies), classics / awsome.

https://archive.org/details/BasicElectricityVol1ToVol5VanValkenburgh/mode/2up

Best wishes
   floor

[George1]

The text below can be found in many of our previous posts. Anyway let us repeat it again for those, who understand simple things with great difficulty.
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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
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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 "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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IF YOU REJECT THE VALIDITY OF THE TEXT ABOVE, THEN YOU REALLY HAVE SOME MENTAL PROBLEMS! AND YOU HAVE TO SEE YOUR DOCTOR!