A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[George1]

Hi Gyula,
Thanks a lot for your reply. Thanks a lot for your brilliant and expert analysis. Please give us some time to consider carefully your last post and prepare the related answers.
Regards,
George
 
   

[George1]

Hi Gyula.
Here are our answers.
1) Yes, you are absolutely right, that it's worth to think over some work-out school examples on hydrogen production. This approach seems to be as if simpler, easier and cheaper. Good idea! We already started discussing it. May be the Hoffman voltameter is most suitable for the purpose.
2) About the Hogen 6m hydrogen generator.
2A) Yes, you guessed correctly that our initial intention to open the electrolyzer cabinet and gain access to the electrodes wiring/cable system etc. was strongly disapproved and denied.
2B) So we have to change the approach. Having in mind the Hogen 6m main technical data (which according to the Hogen's operators coincide with the real measurements) we made some simple calculations.
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A) 40.8 kWh of energy is necessary for the production of 0.6875 kg of hydrogen within a period of 1 hour. Besides 40.8 kWh = 146,880,000 J.
B) LHV of hydrogen (as you wrote in your last post) is 120,000,000 J. Therefore the heat of burning of 0.6875 kg of hydrogen is given by
0.6875 x 120,000,000 = 82,500,000 J.
C) If Hogen 6m is considered as a hydrogen generator only, then its efficiency is given by
82,500,000 J/146,880,000 J = 0.56.
D) Hogen 6m machine operators told us also that AC is converted to DC by a simple standard Graetz rectifier system. Therefore an AC 40.8 kWh of energy is just equal to a DC 40.8 kWh of energy as current flows alternatively through the two "branches" of the Graetz rectifier system 50 or 60 times per second.
E) Let us determine current I (DC) which flows through the electrolyte and through the Graetz rectifier system as (1) the Graetz rectifier system is considered as one united whole and (2) the Graetz rectifier system and the electrolyte are connected in series. The current I is given by
I = (m)/(Z x t)  <=> I = 19000 A
where
m = 0.6875 kg of hydrogen liberated within a period of 1 hour
Z = electrochemical equivalent of hydrogen, kg/C
t = 3600 s
F) The total ohmic resistance R of the connected in series (1) electrolyte and (2) Graetz rectifier system (the latter considered as one united whole) is 0.0001 Ohm approximately.
G) If Hogen 6m machine is considered as a total heat generator, then its COP is given by
(146,880,000 + 82,500,000)/(146,880,000) = 1.56 > 1.
(Notice. 146,880,000 J is the Joule's heat generated by both the electrolyte and the Graetz rectifier system connected in series. Let us remind again that the Graetz rectifier is considered as one united whole whose ohmic resistance R1 is smaller than R, that is, R1 < R or R1 < 0.0001 Ohm.)
H) The Hogen 6m hydrogen generator has systems which keep a constant pure water and cooling agent supply, which on their behalf keep a constant mass, temperature and ohmic resistance of electrolyte and Graetz rectifier. If you touch with your palm the outer surface of the Hogen 6m hydrogen generator, then you feel neither heat nor cold. The temparature is neutral, that is, the temperature is always approximately equal to the temperature of the human body, although the electrolyser has been working without stopping for many hours.
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The above approximate calculations seem to be correct, more or less. They are based on the Hogen 6m's main technical data.
What is your opinion?
Looking forward to your answer.
Regards,
George
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P.S. We intend now to focus all our efforts on some school experiments related to hydrogen generation. Your advice is really good!           

[gyulasun]

Hi George,

Well, it was "expected" they would not let even open the cabinet door of the Hogen 6m machine...
You may have checked the link I gave to the manufacturer's web page on the machine where there is a photo with an opened cabinet door to take a look at the inside parts. You can see it if you scroll down in this link:
https://www.protononsite.com/products-proton-site/h2-h4-h6 
and in the lower right corner you can see the actual Cell Stack which does the electrolysis. Click on the yellow + icons to open further close-ups on the parts.  Because the Cell Stack is very likely a thermally well isolated 'box', no wonder your team could not feel heat to the touch on the side walls of the cabinet, ok.
What you mention under point D)  (what the operators told you) is interesting in that only a Graetz rectifier system is used. One would expect a step-up AC-DC switch mode power supply instead but this is a secondary question of course.

From the spec sheet of the machine https://www.protononsite.com/sites/default/files/2019-02/H%20Series.pdf

it turns out the electrical requirement for the machine is 380-415 VAC, three phase, 50 Hz (or 480 VAC, three phase, 60 Hz).
So suppose we full wave rectify say 400 V, 3 phase 50 Hz AC input and we get say 540 VDC from the Graetz output. 

Now, if 540 VDC is available for electrolysis, then the 19000 Amper current you calculated from the formula would amount to 540*19000*3600 = 36936000000 J (36.936 GJ) energy consumed during 1 hour. 

If we divide 540 VDC by 19000 Amper, the resistance R would be 0.0284 Ohm, this is in conflict with your 0.0001 Ohm estimation. Can you explain this?

because then your COP calculation of 1.56 for the Hogen machine becomes questionable?

How did you arrive at to get 0.0001 Ohm overall resistance for the electrolyte and the rest of the circuit in series with it?

Anyway these are but 'small problems' probably existing in paper only, the focus should really be on doing tests.

Gyula

[George1]

Hi Gyula,
Thanks a lot for your reply.
1) Yes, it seems reasonable to accept 540 VDC from the Graetz output. Please give us some time to re-consider carefully again our test results and, if necessary, to make some additional calculations and carry out some additional tests.
2) Meanwhile, following your good recommendation, we are working hard on some school experiments related to water electrolysis. It will take some time to do everything  in a precise manner.
I will write to you in the nearest future.
Regards,
George

[George1]

Hi Gyula,
We as if managed to clear up the situation.
1) The Hogen 6m machine operators are young men who started working in the factory 3 months ago only. So we contacted one of the older operators who has retired on pension 3 months ago and who told us that Hogen 6m is actually not a very simple machine. The inlet 3-phase AC is reduced to a lower AC and after that converted to DC by a complex and sophisticated system including transformers, diodes, control electronics and other components. So our 540 VDC assumption is obviously not correct.
2) The retired operator likes very much our concept related to considering of any standard water-splitting electrolyzer as a heater of COP > 1. He says that Hogen 6m's electric circuits are too many in number and too complex and sophisticated and it is very difficult to solve the problem related to calculation of each circuit one by one. Instead, he says, it is just enough to use only the figures 40.8 kWh and 0.6875 kg of liberated hydrogen per hour. Because, the operator says, these 40.8 kWh of electric energy transform entirely into 40.8 kWh of Joule's heat no matter what is the resistor (a solid one, a liquid one or a combination of solid and liquid one connected in series (the latter being the Hogen 6m case)) and no matter whether it is an AC or a DC. But in addition to the Joule's heat the liquid resistor generates hydrogen, which if burned/exploded, gives an additional and substantial portion of heat. Our new friend (that same retired Hogen 6m operator) calculates COP as
(146880000+82500000)/146880000 = 1.56 > 1.
So you can see that the former experienced Hogen 6m operator's line of reasoning entirely coincides with ours as we do not try to influence him in any way.
3) About the school and home-made water-splitting experiments.
3A) There are hundreds (and may be thousands) experiments of this kind in Internet, in general, and on YouTube, in partial. All experiments, described in Internet, confirm WITHOUT EXCEPTION our basic concept that any standard hydrogen-generating and water-splitting electrolyser can be considered as a heater of COP > 1. We already repeated tens of times many of these experiments. The easiest ones are with seawater (there is already a big container of seawater in front of our laboratory) and with tap (or pure) water either with table salt (NaCl) or with baking soda (NaHCO3). (The latter being preferred because, if used in electrolysis, NaCl liberates dangerous Cl.)
3B) Now we are preparing an experiment with sulphuric acid. It will take some time, because sulphuric acid is a special and dangerous substance and we have to be very careful. This experiment has to be carried out in a most safety and precise manner.
3C) And one more experimental device is under construction. I will write to you about it in the nearest future.
Looking forward to your answer.
Regards,
George