A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[lancaIV]

Hello George1, http://rexresearch.com/kanarev/kanarev1.htm shows data, measurements and graphs about thermal heatgeneration and water to hydrogen/oxygen dissoziation with propagating efficiencies > 1 !
Partial over 20 years old this information changed not the "scientifical status quo".

Sincerely OCWL

[George1]

To lancaIV/OCWL
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Hi lancaIV/OCWL,
Thanks a lot for your last post.
1) But this is not the same, my friend! And actually this is something entirely different! (Although as if some basic principles coincide -- in both cases it's a matter of electrolysis.) Prof. Kanarev builds expensive, sophisticated and complex devices which on their behalf generate sophisticated and complex electrochemical processes. (The latter are not studied entirely, I am sure, and there are still too many unknown things related to Prof. Kanarev's research.) Our approach is entirely different from Prof. Kanarev's approach. We do not build sophisticated and complex theories. We do not  build expensive, sophisticated and complex devices which on their behalf generate sophisticated and complex electrochemical processes. We simply take a standard ordinary electrolyzer and use it as a heat generator whose efficiency is bigger than 1. Evidently the difference between the two approaches is enormous, isn't it?
2) Anyway your last post is extremely valuable. It shows that in principle it is possible to design and manufacture an electrolysis-based heat generator whose efficiency is bigger than 1.
3) Prof. Kanarev's research is very interesting and two members of our team are studying it very carefully now.
Looking forward to your answer.
Best regards,
George       

[George1]

To gyulasun
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Hi Gyula.
We are doing our best to follow your advices and recommendations.
Firstly, I keep studying hard a heavy textbook (and a few smaller manuals) in experimental calorimetry.
Secondly, in the nearest futute we plan to carry out a few extremely exact and precise calorimetric experiments according to your requirements. (And perhaps the 800-pages report will be ignored as in my poor opinion most of the tests in it are not accurate enough.)
Thirdly, we are searching now for an electrolyzer, which is newer and more reliable than the electrolyzer used in the 800-pages report.
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While preparing ourselves for the above mentioned experiments would you be so polite to have a look at the considerations below? These are as follows.
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1) Let us consider three resistors -- a solid one, a liquid one (an electrolyte) and a gaseous one (atmospheric air for example).
2) Let us apply one and same voltage V=const to each of these three resistors separately.
3) Let us assume that in all three cases we have measured one and same current I=const which flows through each of the three resistors. (The gaseous resistor, i.e. the atmospheric air, generates either a spark or a voltaic/welding arc.)
4) Therefore if the Ohm's law is true, then for any of the above mentioned three resistors are valid the equations
V/I=R (1) <=> V/R=I (2) <=> V=IxR (3)
where R=const is the ohmic resistance of any of the above mentioned three resistors.
5) Therefore if the first Joule's law (related to Joule's heating) and the basic calorimetry laws are valid, then we can write down the following equalities:
E=VxIxt=Q=IxIxRxt=C1xM1x(T1-T)=C2xM2x(T2-T)= C3xM3x(T3-T) (4)
where
t is time/time period;
E=VxIxt is the electric energy generated by the battery of voltage V=const;
Q=IxIxRxt is the heat generated by any of the above three resistors;
C1 is the specific heat of the solid resistor;
M1 is the mass of the solid resistor;
T1 is the temperature of the solid resistor at the end of the time period t;
T is the teperature of any of the above three resistors in the beginning of the time period t;
C2 is the specific heat of the liquid resistor;
M2 is the mass of the liquid resistor;
T2 is the temperature of the liquid resistor at the end of the time period t;
C3 is the specific heat of the gaseous resistor;
M3 is the mass of the gaseous resistor;
T3 is the temperature of the gaseous resistor at the end of the time period t.
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If tested experimentally, all of the above equations have to be true, haven't they?
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Looking forward to your answer.
Best regards,
George

     

[George1]

The last considerations seem to be correct, don't they? Otherwise electric enginnering and calorimetry have to be destroyed. 

[gyulasun]

The last considerations seem to be correct, don't they? Otherwise electric enginnering and calorimetry have to be destroyed.

Hi George,

Well, you start with considerations, from which the 3rd can only be an assumption indeed and may not be correct in practice.
The problem is that in the case of an electrolyte for instance, why do you think the current would be constant?  Because the resistance of the liquid will certainly change as Hidrogen and Oxigen are created and leave from the liquid and also the temperature of the liquid will certainly increase.  Maybe I am wrong but I do not assume liquid resistance hence current during the process remains constant.  Have you or your team found it remaining constant ? Or maybe changing only negligibly ?

This changing current may also be valid for the gaseous 'resistor' and can be a constant indeed for only the 'solid' resistor type.

This then means that the Equations that are made equal to each other while are based on the constant current and resistance assumptions cannot be correct.

If you were to consider an averaged current value for the time duration during which say electrolysis is being done,  then certain Equations would be correct but no need to equate them with each other. 
This would involve either a continuous or a frequently sampled logging of current values from which  an average value can be deduced for the electrolysis, to arrive at the consumed input power hence energy.
It is okay that the voltage would be kept at a constant (stabilized) value.  Here I mention M2 (mass of the liquid) which will be changing (reducing) continuously as the H and O leave from it, have you considered this? 

Hopefully, the specific heat, C2 for the liquid would not change during the electrolysis process, I do not know. 

I mention also that in your test setup described in the paper you started out with,  the gaseous resistor is not needed to consider here in any way, it is irrevelant, no? 

All in all, with the consideration like using the average current with constant DC input voltage,
equation for the input energy taken from the DC supply would be E_in=V x I_average x t

Equation for one part of the output energy, heat, created in the liquid is:  Eout1 = C2 x M2 x (T2-T) 

The other part of the output energy is created by the burning hidrogen, this needs to be decided how you measure it. One possibility is to heat up a given amount of water from t1 to t2 temperature during a measured time duration.
This would involve say using a chamber relatively well isolated from the enviroment so that little heat could escape from inside the chamber as an unmeasurable loss.  Probably there are other, maybe simpler methods. Like for instance to heat up a well insulated room, from say room temperature to a higher temperature, with continuous air mixing inside the room for checking air temperature.  Also,  a good comparison for the amount of heat from the burning Hidrogen would be to use an electric heater in the same room, also starting from the same room temperature and arrive at the same higher temperature and measure the electric input energy of the heater.  This would be a double check on the energy coming from the Hidrogen burning, that is all. 

So the two measured output energies are to be added and their sum then compared to the measured input energy, to get a COP value.   

Gyula