A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[George1]

To those members of good will in this forum, who are searching honestly for the truth.
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No questions as far as I can see -- an indication that everything is clear from our post of January 27, 2021, 03:51:29 PM. This is good. 
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So let us proceed further with our explanations, which we started in our previous post of January 27, 2021, 03:51:29 PM .
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1) Another basic axiom of electric engineering states that all LIQUID conductors (electrolytes) generate a great variety of additional effects (while direct current flows through them, i.e. while electrolysis takes place). Most frequently occurring and most important of these additional effects are as follows.
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1A) Additional effect 1. Transferring of metals from one electrode to the other. Examples of such metals are silver, zinc, copper, aluminium, nickel, chrome, gold, iron, sodium and magnesium.
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1B) Additional effect 2. Generation of chlorine.
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1C) Additional effect 3. Generation of oxygen.
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1D) Additional effect 4. Generation of hydrogen.
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1E) Additional effect 5. Some combination of the above four.
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2) Let us consider additional effect 1. Having in mind our previous post of January 27, 2021, 03:51:29 PM, we can write down that:
A) Each second the standard LIQUID conductor consumes electric energy, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
B) Each second the standard LIQUID conductor generates Joule's heat, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
C) Each second the standard LIQUID conductor transfers copper (or any of the above mentioned metals) from one electrode to the other.
Question: Can we burn/explode in some way the transferred copper in order to generate each second some additional portion of heat?
Answer: No, we can't. Metals are not inflammable under normal conditions.
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3) Let us consider now additional effect 2. Having in mind our previous post January 27, 2021, 03:51:29 PM, we can write down that:
A) Each second the standard LIQUID conductor consumes electric energy, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
B) Each second the standard LIQUID conductor generates Joule's heat, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
C) Each second the standard LIQUID conductor generates a certain amount of chlorine.
Question: Can we burn/explode in some way the generated chlorine in order to generate each second some additional portion of heat?
Answer: No, we can't. Chlorine is aggressive and dangerous for human health, but it is not inflammable under normal conditions.
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4) Let us consider additional effect 3. Having in mind our previous post of January 27, 2021, 03:51:29 PM, we can write down that:
A) Each second the standard LIQUID conductor consumes electric energy, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
B) Each second the standard LIQUID conductor generates Joule's heat, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
C) Each second the standard LIQUID conductor generates a certain amount of oxygen.
Question: Can we burn/explode in some way the generated oxygen in order to generate each second some additional portion of heat?
Answer: No, we can't. Oxygen supports the process of burning, but it is not inflammable by itself under normal conditions.
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5) Let us consider additional effect 4. Having in mind our previous post of January 27, 2021, 03:51:29 PM, we can write down that:
A) Each second the standard LIQUID conductor consumes electric energy, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
B) Each second the standard LIQUID conductor generates Joule's heat, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
C) Each second the standard LIQUID conductor generates a certain amount of hydrogen.
Question: Can we burn/explode in some way the generated hydrogen in order to generate each second some additional portion of heat?
Answer: Yes, we can. And here is the exception. If we burn/explode hydrogen immediately after its releasing, then we could generate each second 0.35J of additional heat.
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5-1) Therefore, having in mind the text above, we can write down that:
A) Each second the standard LIQUID conductor consumes electric energy, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
B) Each second the standard LIQUID conductor generates Joule's heat, which is equal to 1J, that is, (1A) x (1A) x (1Ohm) = 1J/s.
C) Each second the standard LIQUID conductor generates a certain amount of hydrogen.
D) Each second the released hydrogen is burned/exploded immediately after its releasing.
E) Each second the burning/exploding released hydrogen generates 0.35J of additional heat, that is, 0.35J/s.
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5-2) In one word:
5-2A) The inlet electric energy, consumed each second, is equal to 1J/s.
5-2B) The outlet heat, generated each second, is equal to 1J/s + 0.35J/s = 1.35J/s.
5-2C) Efficiency = (1.35J/s.)/(1J/s) = 1.35 <=> Efficiency > 1.
5-2D) The latter is in perfect accordance with our further development of Prof. S. L. Srivastava's solution. (Please refer to our previous posts.)
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Everything seems to be clear now, doesn't it? Please ask questions, if any. We'll be glad to answer.

[George1]

Any comments, suggestions, questions?

[George1]

Still no comments, suggestions, questions?

[George1]

About making a real device. But such a real device is in plain sight of everybody. I will explain this in detail. Please read carefully and thoroughly the text below.
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1) Have a look at your car's battery. What happens while you recharge it?
1A) Firstly, you generate Joule's heat. (Please refer to our previous posts.)
1B) Secondly, you generate hydrogen. If this hydrogen is burned/exploded, then a second portion of heat would be generated. (Please refer to our previous posts.)
1C) Thirdly, you store electric energy when the batterry is completely charged. If you connect the already completely charged battery to an ordinary Ohmic resistor (thus forming a circuit) and then if you discharge the battery, then a third portion of heat (this time again Joule's heat as in item 1A) would be generated.
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2) In one word, we can write down that
(B + C)/(A) = 1.35 <=> (B + C + D)/(A) > 1.35 <=> Efficiency > 1.35
where
A = electric energy, which is consumed for recharging
B = first portion of heat, mentioned in item 1A
C = second portion of heat, mentioned in item 1B
D = third portion of heat, mentioned in item 1C
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Simple, clear and understandable. Please refer again to our previous posts, if necessary.
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3) As mentioned many times in our previous posts our DC electrolysis OU concept (which is our second technology breakthrough) is absolutely free.
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4) Our third technology breakthrough however is not free. Our third technology breakthrough is for sale. Let us remind again that our third technology breakthrough increases many times (twice as a minimum and more than 15 times as a maximum) the capacity of any standard now-existing battery. Our third technology breakthrough has both a theory and a working prototype.
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5) Besides we (our team) have designed another 7 technology breakthrough inventions, which are ready for selling too. (We are working now over our 11th invention.)

[George1]

Any comments, suggestions, questions, which are related to our last posts?