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Author Topic: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1  (Read 34311 times)

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #75 on: March 16, 2019, 11:35:39 AM »
To F6FLT.
To lancaIV.
------------------
Hi guys.
1) Please excuse me, if I have offended you in some way. I am sorry for this.
2) We are starting carrying out the real experiments. Any good ideas for simple and reliable experimental step-by-step procedures?
Looking forward to your answers.
Best regards,
George

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Offline gyulasun

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #76 on: March 17, 2019, 04:07:03 PM »

Hi George,

 I do not wish to tell you how to proceed with the actual tests. We have discussed several times what are to be measured, at least I wrote about them. How you achieve them is your solution, and whatever results you get, please report them.  And if I or anyone else here asks questions on the measuring methods and devices you eventually used, then you can hopefully give answers, with photos on the setup etc. No need for a 800 page long report either.

I agree the biggest problem is to collect the deliberated hydrogen and then burning it so that from the heat created only a minimum quantity could escape into the enviroment and much part of the heat should heat up a given quantity of (say) water (or oil).

Perhaps local labs at universities or colleges or at high schools can give some equipment in this respect or they let them use by you in their lab. You do not have to tell them what exactly you want to prove, just say that you wish to perform an electrolysis with correct measurements that include the liquid's temperatures and the performed work of the burning hydrogen on heating up another liquid.  Maybe they have a ready chamber for this latter process.  Or the chemics or physics teacher can advise you on cheap possibilities.  Perhaps start with figuring out in advance the quatity of water for instance, how much heat is needed to raise say half a liter of water from room temp to say 50 degree Celsius and whether this could be done in a heat isolated chamber from which only a minimum amount of heat could escape. etc etc.

Gyula

Offline Floor

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #77 on: March 21, 2019, 09:17:54 PM »
--------------------
Therefore the link above actually explains everything. 
In order to maintain M2=const, T2=const, I=const and R=const in the electrolyte you have to do only two things.
Firstly, you have to add constantly only pure water (as H2SO4 is not consumed in the reaction as shown in the above link and in the above quote) in the electrolyzer thus keeping M2=const.
Secondly, you have to cool down constantly the electrolyzer thus (a) consuming constantly the Joule's heat for useful purposes and (b) keeping T2=const, I=const and R=const. (Because as you know the ohmic resistance of any electrolyte depends on temperature, that is, the ohmic resistance of any electrolyte decreases with rise in temperature. In order to avoid this you cool constantly the electrolyte thus keeping constant values for T2, I and R, respectively.)
----------------------

Change in the temperature effects resistance, as resistance changes and applied voltage is constant, electric
current varies.

Measuring  input wattage not current (use a watt meter),  will simplify / eliminate this aspect  / Question of
temperature variation of the electrolyte over time / need to maintain a constant current over the time duration.

               floor

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #77 on: March 21, 2019, 09:17:54 PM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #78 on: March 26, 2019, 10:39:21 AM »
Hi guys.
Let me report what we have done until now.
We attack vigorously the problem, related to the required real experiment. We found a HOGEN H6m hydrogen generator at a distance of 100 km from the place we live. Every day at least two members of our team travel and cover this distance of 100 km in order to carry out a set of experiments. It will take some time. But we will do it!
--------------------------
Meanwhile we came upon some very interesting things.
Please have a look 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.
--------------------------
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.)
--------------------------
The above solved problem has a potential which can be developed further. And here it is.
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.
------------------------------
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.
-----------------------------
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.
-----------------------------
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!
-----------------------------
Looking forward to your answers.
Regards,
George       

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Offline gyulasun

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #79 on: March 27, 2019, 11:42:04 PM »
Hi George,

I thought you would arrive at some work-out school examples on hydrogen production.   8)   No problem with this approach because from such calculations one can estimate in advance what hydrogen quantity could be expected to receive from a (hopefully also) known input power.
So the calculations show that one could achieve a COP of 1.35 or 1.37. Now the question is how this COP number comes out in practice ?
I think you need to start with either a DC or AC source available and work from there. If an AC source is used for electrolysis, then there is a certain conversion efficiency involved for the AC to DC converter. I may sound as if I am kidding with such details but I am not: you and your team will surely face this when examining the Hogen H6m hydrogen generator (i.e. the Series H from manufacturer Proton Onsite Electrolyzers) in this respect.
Even though it is a professionally 'sounding' generator, its efficiency is written in a book as 50.6 % + 10 % i.e. around 61 %. The efficiency for the Series C (from the same manufacturer) is 59 % + 10 % = 69 %, this indicated by the book as the highest efficiency product among their hydrogen generator family. The 10 % addition is the energy removed earlier from the overall system efficiency so I added them up. The reason is the hydrogen should be dried to comply with the required purity specifications. Drying needs additional energy (about 10%) from the AC mains input and the liberated hydrogen goes through the built-in dryer. 

The book in which I found these data can be read online, see "Chapter 3.2.3.3 Proton Onsite PEM Electrolyzer" here (pages 136 and 137 and PEM is short for Proton Exchange Membrane):

https://books.google.com/books?id=dyEtAgAAQBAJ&pg=PA136&dq=HOGEN+H6m+hydrogen+generator&hl=en&sa=X&ved=0ahUKEwjzxKK8wqDhAhUGt4sKHa5LA70Q6AEIJzAA#v=onepage&q=HOGEN%20H6m%20hydrogen%20generator&f=false

(If  Page 137 comes up blank, try to scroll down a few pages and then back, it will become visible)

and here is the manufacturer site on their Series H machines:
https://www.protononsite.com/products-proton-site/h2-h4-h6   There is PDF file with the specifications for the H6 machine.

Maybe your team members travelling to the place will be allowed to measure the average DC current and DC voltage which actually does the electrolysis ? Unfortunately, to do this (and supposing it will be permitted), the electrolyser cabinet should be opened to gain access to the electrodes wiring/cable system etc. 

Notice 1. You used the HHV data of hydrogen which is ok when you utilize the latent heat of vaporization too that appears say as "hot air" (if I am not mistaken) as the result of the hydrogen gas burning with the ambient air oxigen while the flame heats up say a bucket of water. For the shake of completeness, I would consider the lower heating value, the LHV of the hydrogen too, which is 119.96 MJ/kg and in your 1st book example the outlet energy 2 in this case would be 11996 J. So the COP in your calculated example would be (37446+11996)/38232=1.29 this is no problem for you because still above 100%.   :)

Notice 2. You wrote: "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."

Well, the latter reasoning may sound logical but the question is whether you accept the efficiency specified for the H6m type hydrogen generator as 61 % or you are ready to check the average current and voltage the machine actually uses during a chosen time duration and then estimate input energy from those measurements? Provided of course whether such measurements are allowed by the owners or operators of the machine,
Knowing the actual input energy would greatly help estimating COP and would avoid the AC-DC conversion and other extra losses involved with the machine, provided the exact amount of hydrogen is correctly measured by the machine under a chosen time duration what the machine surely does, no doubt. 

Notice 3. You wrote: "Therefore the text above unambiguously shows that it is a matter of exact experimental data which is in perfect accordance with theory."

Well, I cannot disconfirm whether the numbers used in the two calculation examples you took from the books are obtained by actual measurements, I 'have to accept' they are practically close to reality. I may 'have to accept' also that the 'highly qualified experts' actually measured the input power for instance by monitoring the input current and voltage and I 'have to believe' that this measured power then corresponded to the calculated 31.86 W (or the 37 W),  we simply 'have to' accept this. This is not nit-picking from me, just a notice that you still do not have correct measurements results.

Hopefully, you and the team get closer and closer to obtain real and measured data. I am not against your claims.

Gyula

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #79 on: March 27, 2019, 11:42:04 PM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #80 on: March 30, 2019, 02:08:42 PM »
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
 
   

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #81 on: March 31, 2019, 05:19:45 PM »
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.
---------------------
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.
-----------------------
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
---------------
P.S. We intend now to focus all our efforts on some school experiments related to hydrogen generation. Your advice is really good!           

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #81 on: March 31, 2019, 05:19:45 PM »
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Offline gyulasun

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #82 on: April 02, 2019, 03:28:50 PM »
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

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #83 on: April 08, 2019, 03:39:35 PM »
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

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #83 on: April 08, 2019, 03:39:35 PM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #84 on: April 09, 2019, 01:33:46 PM »
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

           

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #85 on: April 10, 2019, 10:56:59 AM »
Hi again Gyula,
Here is my next report.
The retired former Hogen 6m's operator became a strong supporter of our cause. He considered carefully all posts in this topic written until now and noticed something that had to be corrected. Our new friend and supporter shares the fundamental point of view that the validity of the Joule's heat law directly derives from the Ohm's law and vice versa, that is,
V = I x R  (1)  <=>  V x I x t = I x I x R x t  (2)
where
V is the voltage of the battery;
I is the current flowing through the resistor;
R is the ohmic resistance of the resistor;
t is time.
(Note. Both sides of equality (1) are simply multiplied simultaneously by (I x t) and the result is equality (2).)
But, as our new friend notices, the last two equalities (1) and (2) are strictly valid only for solid resistors. For liquid resistors (electrolytes) equalities (1) and (2) have to be re-written again in a little different manner, that is,
(V - v) = (I - i) x R  (3)  <=>   (V - v) x (I - i) x t= (I - i) x (I - i) x R x t  (4)
where
v is the minimum voltage necessary for the water-splitting electrolysis to begin; v = 1.5 volts by definition;
i is the related small decreasing of current I, caused by the presence of v.
But if V is much bigger than v (that is, if for example V = 100 volts and v = 1.5 volts), then we can assume that equalities (1) and (2) are perfectly valid for the liquid resistor (electrolyte) too as v and i can be neglected.
And from here follows again the expression for COP, which is given by
COP = ((V x I x t) + (H))/(I x I x R x t) > 1  (5)
where H is the heat of burning/exploding of hydrogen generated, HHV or LHV. 
---------------------
So the fact that a highly-qualified and experienced man of more than 30 years of practice related to electrolysers became our supporter gives us an additional strong confidence that the water-splitting electrolysis is really a heating process of COP > 1. It simply follows from (1), (2), (3), (4) and (5).
Looking forward to your answer.
Regards,
George
---------------------
P. S. We keep carrying out experiments. All tests until now confirm the validity of (1), (2), (3), (4) and (5).
   

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #85 on: April 10, 2019, 10:56:59 AM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #86 on: April 11, 2019, 12:48:52 PM »
Hi Gyula,
Let me report what we have done.
1) Our (already) numerous experiments show that the theoretical value of the minimum potential difference of 1.484 volts, which is necessary for the water-splitting electrolysis to begin (Prof. S. L. Srivastava, M. Sc., Ph. D., Solved Problems in Physics, Volume - 2, solved problem 12.94, p. 875), is actually a little bigger and varies between 1.7 and 2 volts. Obviously this is due to the electrode potential, overvoltage, side reactions, etc. But this 1.7 - 2 volts correction practically does not influence the validity of the main concept, i.e., the validity of the inequality COP > 1.
2) Besides there are at least 10 (ten) extremely precise and detailed experimental research papers written by a bunch of highly qualified electrochemistry experts from Japan, India and China. (These three countries seem to the leaders in the field of water-splitting electrolysis as a theory and practice.) We simply took their experimental results. The new summation result was again COP > 1. (It is important to stress upon the fact that the experimental data from the above mentioned research papers is obtained in a much more precise manner than our one -- this is due to the presence of high-quality equipment and highly-qualified experimenters.)
3) Our new friend and supporter, the retired Hogen 6m's operator, says that our COP > 1 conception simply gathers together TRUE experimental facts, which have been WRONGLY considered IN ISOLATION until now.
4) We keep carrying out experiments.
Looking forward to your answer.
Regards,
George

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #87 on: April 15, 2019, 11:24:56 AM »
Hi Gyula,
I am sending to you the sulphuric acid experimental results as promised. We simply repeated the experiment described in solved problem 12.97 (Solved Problems in Physics", 2004, Volume 2, p. 876, Prof. S. L. Srivastava, Ph.D.). The experimental approach and the test results are briefly described below.
1) A glass container, which has a form of a rectangular parallelepiped and which has dimensions 0.01m/0.01m/0.37m, is filled entirely with 30 % sulphuric acid. Two electrodes are dipped in the electrolyte in the two opposite ends of the container. The ohmic resistance of the electrolyte is just equal to 0.5 Ohm. The glass container has no upper lid -- it is open from above.
2) The glass container is situated on a horizontal table in such a manner that its longest side (0.37m) is horizontal,i.e., parallel to the horizontal Earth's surface.
3) The electrolyte is connected in series to a variable resistor (rheostat) in order to control and adjust (if necessary) the value of the current. And more precisely, the circuit consists of a DC source, a variable resistor (rheostat) and a glass container filled with sulphuric acid. These three components are connected in series.
4) Within a period of 20 minutes it can be clearly observed that one of the electrodes generates bubbles of hydrogen (the latter produces flame/explodes slightly if fired) and the other electrode generates bubbles of oxygen.
5) In order to keep a constant value of the current without using the rheostat we had to keep pouring (from time to time) pure water in the container and keep cooling it down. (It was not an easy operation and was a little dangerous.)
6) The experimental COP results always varied around 1.29 (which was calculated by you assuming that hydrogen's LHV is equal to 120 MJ per kilogram of liberated hydrogen). Sometimes we got COP = 1.21, sometimes COP = 1.37, sometimes COP = 1.28, etc.; the mean value being around 1.29.
7) We have carried out already almost 100 experiments using various electrolytes in order to split water into hydrogen and oxygen. COP was always bigger than 1, i. e., COP > 1. Most of the already carried out experiments can be found in Internet and on YouTube and we simply coppied/repeated them.
8) It seems to us that there is no sense to keep carrying out other experiments. Their number is already equal to 100 and all these 100 experiments unambiguously show that experimental results confirm theory. COP > 1.
9) Besides (as mentioned in our previous post) there are at least 10 (ten) extremely precise and detailed experimental research papers written by a bunch of highly qualified electrochemistry experts from Japan, India and China. (These three countries seem to the leaders in the field of water-splitting electrolysis as a theory and practice.) We simply took their experimental results. The new summation result was again COP > 1. (It is important to stress upon the fact that the experimental data from the above mentioned research papers is obtained in a much more precise manner than our one -- this is due to the presence of high-quality equipment and highly-qualified experimenters.)
Looking forward to your answer.
Regards,
George           

Offline gyulasun

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #88 on: April 15, 2019, 11:42:28 AM »
Hi George,

Thanks for your posts on the activity you and the team have been doing on this electric heater topic.

It is good that you can get help from the now retired former Hogan 6m type machine operator.
It is even better that you found out that textbook examples cannot be trusted 100% in practice (electrode potential issue etc what was already mentioned by F6FLT), even if Professor M.Sc. Ph.D persons write them for students.

Regarding the research papers you mention, well, they may have more scientific approach and test results than textbook examples but of course you need to take and interpret them carefully too. If you do not mind I would be interested in the papers titles and their authors, just out of curiosity.

All in all, the only thing to achieve is to prove the COP > 1 claim for your setup by correct measurements if you want scientific community accept the claim. Especially so when such a setup is to be marketed as a product, having an unusually high efficiency that beats any other heaters already in use.

PS  I already wrote this answer in Notebook when I noticed your latest answer a few minutes ago, will return later.

Gyula

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #89 on: April 15, 2019, 12:48:16 PM »
Hi Gyula,
Thanks a lot for your reply.
Looking forward to your answer related to my last post.
Regards,
George

 

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