A simple electric heater, which has efficiency greater than 1

[Leafy]

Solution. The power consumed is equals to 31.86W.

Notice that the power consumed is 31.86W. There is no saying that is the heat generated.

[PeterAX]

To Leafy.
---------------------------------
My name is Peter. I am a member of George's team. George is now fully occupied with some extremely precise experiments and with some extremely sophisticated R&D work. So he asks me to keep the correspondence related to this topic.
----------------------------------
A few words about your last post.
---------------------------------
1) One of the basic axioms of electric engineering is given by the equality
P = I x I x R = Q,
where
P = electric energy, which is consumed each second by the conductor
I = direct current, which flows through the conductor
R = Ohmic resistance of the conductor
Q = Joule's heat, which is generated each second by the conductor
-------------------------------------
2) Equality P = I x I x R = Q is valid for any standard conductor (no matter solid, liquid or gaseous).
-------------------------------------
3) Equality P = I x I x R = Q can be found in any electric engineering beginner's guide/manual/textbook.
-------------------------------------
4) Prof. S. L. Srivastava simply solves the problem by using the above mentioned axiom of electric engineering. (The same for his two Russian colleagues some 40 years ago.)
--------------------------------------
Everything seems to be clear now, doesn't it?

[Tarsier79]

Whack-a-mole. Hi George. You still aren't fooling anyone.

[WaltzCee]

4) Prof. S. L. Srivastava simply solves the problem by using the above mentioned axiom
of electric engineering. (The same for his two Russian colleagues some 40 years ago.)

40 years they've been on to this and no practical use.
We're just not buying it George.

[Leafy]

-------------------------------------
2) Equality P = I x I x R = Q is valid for any standard conductor (no matter solid, liquid or gaseous).
-------------------------------------

I find this hard to convince especially when the conductor is making hydrogen.

[WaltzCee]

You fine individuals are pissing up the wrong tree. I doubt I can point you to the right one
but let me give it a try.

National Renewable Energy Laboratory
1617 Cole Boulevard, Golden, Colorado 80401-3393
303-275-3000 • www.nrel.gov
Operated for the U.S. Department of Energy
Office of Energy Efficiency and Renewable Energy
by Midwest Research Institute • Battelle

Talk to those people.

Oct 7, 2019In the HD scenario, U.S. hydrogen supply becomes a $130 billion
per year industry by 2050 (Fig 2). Hydrogen sector revenues and profits rise rapidly through
the early 2030s, with revenue growth...

If you can capture the heat, convert it to electricity then use that for further hydrogen
production you'll be miles ahead of any commercially available units. You'll put everyone
else out of business! All we'll get out of it is we'll be able to say we knew you when you
were just a bunch of annoying dickweeds.

[PeterAX]

To WaltzCee.
---------------------------
Thank you for your reply.
----------------------------
You wrote:"40 years they've been on to this and no practical use."
---------------------------
1) Fisrtly, Prof. S. L. Srivastava (and his two Russian colleagues some 40 years ago) only solved a standard problem in a standard manner. We (our team) developed further Prof. S. L. Srivastava's solved problem in a non-standard manner. Our further non-standard development of Prof. S. L. Srivastava's solved problem led to efficiency > 1. Please read carefully one of our next posts, which has been published already many times here.
---------------------------
You wrote:"We're just not buying it George."
---------------------------
2) Secondly, I am not selling anything. Our water-splitting electrolysis OU concept is absolutely free. And as if this fact is more than evident. We need only public recognition.
--------------------------
3) Our third technology breakthrough however is not free. Our third technology breakthrough is for sale.

[PeterAX]

To WaltzCee.
--------------------------------
You sent to me the contact information below.
--------------------------------
National Renewable Energy Laboratory
1617 Cole Boulevard, Golden, Colorado 80401-3393
303-275-3000 • www.nrel.gov
Operated for the U.S. Department of Energy
Office of Energy Efficiency and Renewable Energy
by Midwest Research Institute • Battelle
--------------------------------
Thank you very much for the the contact information above. Much obliged!

[PeterAX]

To WaltzCee.
---------------------------
My respect to your creative and non-standard way of thinking! You just gave an excellent idea!
---------------------------
You wrote:"If you can capture the heat, convert it to electricity then use that for further hydrogen production you'll be miles ahead of any commercially available units."
Yes, to capture the heat! Perfect! You developed further our team's concept! Any theoretical (only theoretical) idea how to capture the heat?

[PeterAX]

To WaltzCee.
----------------------------------
The text below can be found in many of our previous posts. Anyway let us repeat it again.
-----------------------------
Have a look again 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=rrKFzL ... 22&f=false
--------------------------
For your convenience I am giving below the text of the problem and its solution.
--------------------------
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.)
--------------------------
WE DEVELOPED FURTHER PROF. SRIVASTAVA'S SOLVED PROBLEM IN A NON-STANDARD MANNER.
OUR FURTHER DEVELOPMENT OF PROF. SRIVASTAVA'S SOLVED PROBLEM LED TO COP > 1.
HERE IS THE ESSENCE OF OUR APPROACH.
--------------------------
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.
------------------------------
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.
-----------------------------
And one more interesting fact.
Literally the same solved problem can be found in an old Russian (still from the Soviet times) book "&#1057;&#1073;&#1086;&#1088;&#1085;&#1080;&#1082; &#1079;&#1072;&#1076;&#1072;&#1095; &#1080; &#1074;&#1086;&#1087;&#1088;&#1086;&#1089;&#1086;&#1074; &#1087;&#1086; &#1092;&#1080;&#1079;&#1080;&#1082;&#1077;", 1986, p. 130, solved example problem 71. The authors of this book are &#1056;. &#1040;. &#1043;&#1083;&#1072;&#1076;&#1082;&#1086;&#1074;&#1072; and &#1053;. &#1048;. &#1050;&#1091;&#1090;&#1080;&#1083;&#1086;&#1074;&#1089;&#1082;&#1072;&#1103;. 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.

[PeterAX]

To Leafy.
--------------------------------
You wrote: "I find this hard to convince especially when the conductor is making hydrogen."
-------------------------------
Well, imagine that there are two closed identical boxes. In one of the boxes is hidden a solid conductor and in the other of the boxes is hidden a liquid conductor. Each of the two conductors is connected to a standard DC source thus forming a circuit. Each circuit is equipped with an ammeter and with an ohmmeter, that is, we have two identical DC sources, two identical ammeters and two identical ohmmeters. The two identical ammeters have two identical readings. The same for the ohmmeters -- the two identical ohmmeters have two identical readings.
Question: How can we guess in which box is hidden the solid conductor and in which box is hidden the liquid conductor?
Answer: No way to guess.
Do you agree with the text above?

[Leafy]

Yes, I do agree. But what you measured is just the input. How are you going to measure the output? I can just put a battery to charge in the box and you wouldn’t know.

[WaltzCee]

I didn't think it would do any good, George.

50,755 words and all it shows is you are all talk. If you had a viable idea you would just do
it.

[PeterAX]

To Leafy.
-------------------------------------
You are not reading carefully my posts.
-------------------------------------
1) One of the basic axioms of electric engineering is given by the equality
A = I x I x R = B,
where
A = electric energy, which is consumed each second by the conductor
I = direct current, which flows through the conductor
R = Ohmic resistance of the conductor
B = Joule's heat, which is generated each second by the conductor
-------------------------------------
2) Equality A = I x I x R = B is valid for any standard conductor (no matter solid, liquid or gaseous).
-------------------------------------
3) Equality A = I x I x R = B can be found in any electric engineering beginner's guide/manual/textbook.
-------------------------------------
4) Please refer to any electric engineering beginner's guide/manual/textbook and see how A and B are calculated.
(Please do not force me to read you a lecture on basic axioms of electric engineering, dear colleague! You are a technology expert, aren't you?)

[WaltzCee]

Question: How can we guess in which box is hidden the solid conductor and in which box is hidden the liquid conductor?
Answer: No way to guess.
Do you agree with the text above?

Wrong answer. If you want to hire an engineer, go to indeed. If you want to be an engineer,
read a book.