A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[George1]

To F6FLT.
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You greatly surprise me, my friend! You are an expert in mechanics as well as in electric engineering! (And may be in any other field of technology?)
You have written: "  You have simply forgotten the oxidation-reduction potentials! The potential difference that will result in heating is therefore less than the one used. This means in other words that even for the same current, the energy used for producing hydrogen is not used to heat." There is no sense in this composition of words. This is for example something like the following sentence: " The Moon is black and it walks around the green tree." Grammatically correct, but absurd.

[George1]

Dear colleagues,
My name is George Sen. I am a member of a team of inventors. Please have a look at the link
https://mypicxbg.files.wordpress.com/2019/01/pages_1-6.pdf
The link above describes a simple electric heater, which has efficiency greater than 1.
What do you think about this electric heater? What is your opinion?
Looking forward to your answer.
Best regards,
George

[lancaIV]

Only"Q"~ heater efficiency nearly 100% ,  "Q+H" ~ electrolyzer, catalyzer : efficiency over 100%
A. How many parts from "H": 33 KWh are double calculated from the "Q": 50 KWh  ?
a1: only calculation or physical measured ?

When not only "resistive heater"= pure Joule change then catalytic Joule process possibility :
https://www.google.com/search?q=hasebe+hydrogen+patent&ie=utf-8&oe=utf-8&client=firefox-b
Sincerely
OCWL

[George1]

Hi lancaIV/OCWL,
Thanks a lot for your reply.
1) Actually I did not understand some parts of your text. Would you be so polite to make some parts of your message a little clearer?
2) Electric energy E=VxIxt (which is generated by the battery) transforms ENTIRELY into Joule's heat (which is generated by the resistor) as Joule's heat=Q=IxIxRxt. And this fact is valid for a solid resistor as well as for a liquid resistor. In other words, (a) if you put a resistor into a closed box and (2) if your voltmeter and ammeter show that V=const and I=const, then you will not be able to guess whether the resistor inside the box is solid or liquid. Ohm's law and Joule's first law are valid for any solid resistor as well as for any liquid resistor. However a liquid resistor like the electrolyte, used for electrolysis of water, generates hydrogen in addition. 
3) Physically measured. At the inlet we measured V, I and t by using a voltmeter, an ammeter and a chronometer, respectively. At the outlet we carried out ENTIRELY CALORIMETRIC experiments by measuring (a) the heat generated by the electrolyte and (b) the heat generated by the burning of hydrogen. We used a standard calorimeter -- nothing special.   
Looking forward to your answer.
Best regards,
George   

[gyulasun]

Hi George,

You have an interesting idea of heating up electrolyte by DC input and utilize the heat and also utilize the heat from the burning Hidrogen received from the electrolysis too. 

Would like to ask that from the tests what was the result? What efficiency numbers did you find which were consistently higher than 1? How much uncertainty do you think may have occured when checking the heat quantity the Hidrogen provided? What method did you follow for estimating it?  (The heat developed in the electrolyte is easy to measure by a calorimeter of course.)

You did not mention Oxigen in the paper while it is also created during the electrolysis process, I suppose.  Or you found that burning only the created Hidrogen already pushes overall efficiency > 1 ?
I would also be curious about the DC current level used for heating the electrolyte.   

Thanks,  (I know I have many questions...)   
Gyula