To F6FLT.
To lancaIV.
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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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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.)
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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

Hi George,

I thought you would arrive at some work-out school examples on hydrogen production.      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

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!           

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

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. 
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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
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P. S. We keep carrying out experiments. All tests until now confirm the validity of (1), (2), (3), (4) and (5).
   

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.
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           

Hi Gyula,
Thanks a lot for your reply.
Looking forward to your answer related to my last post.
Regards,
George