A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[George1]

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

[gyulasun]

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

[Floor]

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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

[George1]

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!
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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
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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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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.
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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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Looking forward to your answers.
Regards,
George       

[gyulasun]

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