A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1

[tinu]

Hi tiny,
But you are making an obvious mistake. The generated power is equal to 0.77 W and the consumed power is equal to 0.77 W too. Actually "the countervervoltage" (which is due to the electrode potential/overvoltage) plays the role of small battery that is connected in opposite direction to the main DC source, i.e., plus to plus and minus to minus. Imagine that the main DC source has a voltage of + 1,23 V. Then what will be the value of the consumed/generated power? Answer: just equal to zero.
...

Hi George1,

In case you're talking to me, please note I'm not tiny. For now, just take my word for it.
Other than that, please go do the simple experiment above. Most members here can do it for breakfast and still have some time left. I hope you are able to do it too. If not, many members can help you.

Last, but not least, please stop polluting us with utter nonsense like that right above and elsewhere. A standard liquid resistor is not equivalent with an electrolysis cell! An electrolysis cell involves a standard liquid resistor PLUS the standard potential for dissociation. If you're not able to grasp these very simple theoretical concepts, maybe a 10$/15min experiment may get you out of the state of confusion you're in right now?

Best regards!

[George1]

Hi tinu,
You are trying to manipulate all of us here in this forum -- you are trying to convince us that black is white. But you are not a skillful manipulator. You have to read still more books related to the art of manipulation. I will not argue with you.
George

[tinu]

Hi tinu,
You are trying to manipulate all of us here in this forum -- you are trying to convince us that black is white. But you are not a skillful manipulator. You have to read still more books related to the art of manipulation. I will not argue with you.
George

Hi George1,

Now you start being rude, don't you think?
I'm purely stating elementary knowledge, backed up by references you can understand (wiki) and by experiments. Some of these experiments I've done by myself more than 30-35years ago as part of the learning process. Electrolysis too. Since then, I've repeated them many times with a twist. Regarding electrolysis in particular, there are several good threads into this forum you might want to study. Novel ideas, interesting approaches but ... no cigar so far. In short, you don't need to reinvent the wheel now for the sake of calling yourself an inventor. Or if you want it so badly as to satisfy your ego or to fulfill a hidden agenda you might have, please do it in a way that does not require propagating blatant untrue statements. 

And please stay to the facts, no ad-hominem attacks. Post references and/or experimental results if you have any.
Help people progress in the field if you can, instead of throwing some into dead ends by misleading them. If you cannot help, please refrain yourself, ask questions and keep on learning. This is a public forum available without restriction so I'll not allow any person (you included) to shout worldwide ever so often (almost daily in your case?!) how magnificent his/her idea are when, in fact, they are unfortunately but truly flawed. Ok? And no offense, but you reveal a level of education in physics of an undergraduate student. This is not an insult by any means (I'm also very uneducated in most of the fields except very few) but it's stated for the reason to kindly ask you to show some respect to the experts here and elsewhere as well as to every visitor coming to read us. I have a M.Sc. in physics. I understand you are an enthusiast but what is exactly your level of expertise?

You are welcome to argue with me.

Best regards!

[George1]

Hi tinu,
First of all please excuse me if I have insulted you in some way. I am really sorry about this.
=================
Ok, let us follow your rules.
------------------------------
Given:
V = 2 Volts
v = - 1.23 Volts
R = 1 Ohm
I = (V - v)/R = (2 - 1.23)/1 = 0.77 Ampere
t = 1 second
Z = 0.00000001044 kg/C
Hydrogen HHV = 142000000 J/kg
Hydrogen LHV = 120000000 J/kg
-----------------------------
inlet energy = (2 Volts) x (0.77 Ampere) x (1 second) = 1.54 Joules
-----------------------------
outlet energy 1 = (0.77 Ampere) x (0.77 Ampere) x (1 Ohm) x (1 second) = 0.5929 Joules
-----------------------------
outlet energy 2 = (0.00000001044 kg/C) x (0.77 Ampere) x (1 second) x (142000000 J/kg) = 1.1415 Joules
-----------------------------
COP = (outet energy 1 + outlet energy 2)/inlet energy = (1.1415 Joules + 0.5929 Joules)/1.54 Joules = 1.1262  <=>
<=> COP = 1.1262  <=>  COP > 1.
-----------------------------
In the above solution we use hydrogen's HHV. If we use hydrogen's LHV, then COP will be equal to 1.01, that is, COP = 1.01  <=> COP > 1.
Of course it is aways better to use hydrogen's HHV instead of hydrogen's LHV. We have a choice.
-----------------------------
So you see that COP again is bigger than 1, that is, we have again COP > 1.
-----------------------------
Looking forward to your answer.
George


     

[tinu]

 Hi George1,

Apologies accepted.

Now to the subject, I said let's have 2V and 1A . You start there but immediately later you change it to 2V and 0.77A by the improper use of Ohm law. That's not correct. Let's do it right, ok? Here it goes:

Electrolysis require a potential drop of min. 1.23v (practically it's larger, depending on many factors). This is a forward drop of voltage, not a counter emf. It's a forward voltage drop similar (for the purpose of simple understanding only) to the forward drop of about 0.6V on a conducting diode (p-n junction). I repeat: there is no counter voltage in electrolysis!

The proper use of Ohm law and Kirchhoff's voltage law is:
Vps=IpsxR + Ve,
where:
Vps is the voltage applied to the electrolysis cell by the power supply (ps) that can be measured with a voltmeter, in our case Vps=2V
Ips is the current supplied to the electrolysis cell by the power supply (ps) that can be measured with am ammemeter, in our case Ips=1A
Ve=1.23V is the potential difference required for electrolysis to occur.
R is the resistance of the electrolyte (the equivalent resistance of liquid resistor) in a very simplified model.

In the above case, you can use Ohm law and compute R = 0.77 Ohm but this value is of little practical use because this model is over-simplified while a real electrolytic cell involves higher complexity.

Regardless on complexity, in our model we have:
Supplied power is Vps x Ips, as measured by the two multimeters, 2V x 1A = 2W
Meanwhile, Joule power is given by sqr(Ips) x R, in our case sqr(1A) x 0.77Ohm = 0.77W
In the same time, electrolysis requires Ve x Ips which is 1.23V x 1A = 1.23W

Part of the supplied power goes into heat (0.77/2=38.5%) while electrolysis efficiency in this case would be 1.23/2=61.5%. Overall COP=1.
Besides theory and correct use of the laws of electricity, you may also easily perform this simple experiment.

In good quality electrolytic cells, efficiency can go toward 80%, meaning that only about 20% of total energy supplied is lost, mainly as Joule heat. In fact, I saw you did reference a fine paper (http://www.esru.strath.ac.uk/Documents/MSc_2003/papagiannakis_i.pdf) where energy efficiency of electrolysis was measured between 73-84% (page 72).

I'm sorry but there is no overunity here.

Best regards!