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Author Topic: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1  (Read 168316 times)

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #510 on: February 18, 2021, 02:56:04 PM »
A hot discussion in besslerwheel.com/forum. The same topic and the same title. Our opponents totally lost psychic balance. They try to reject the Joule's heat formula, which is one of the basic axioms of electric engineering! This is really a shocking absurd!

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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #511 on: February 19, 2021, 03:58:03 PM »
Any comments, questions, recommendations?

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #512 on: February 22, 2021, 02:23:40 PM »
Ok, let us recapitulate our many-months correspondence and summarize it in one single text. And this text is given below.
-----------------------------------------------------------------------
1) Let us connect a standard DC source to a standard conductor thus forming a circuit.
--------------------------------------
2) We can write down the following two related equalities
V = I x R (1)  <=>  V x I x t = I x I x R x t (2),
where
V = voltage of the DC source
I = direct current, which flows through the conductor
R = Ohmic resistance of the conductor
t = time period, within which direct current flows through the conductor
----------------------------------------
3) Equality (1) is the mathematical expression of the Ohm's law.
-----------------------------------------
4) Equality (2) is the mathematical expression of the Joule's law of heating. 
----------------------------------------
5) Equality (2) can be derived from equality (1) by multiplying both sides of (1) by (I x t).
-----------------------------------------
6) And vice versa, equality (1) can be derived from equality (2) by dividing both sides of (2) by (I x t).
-----------------------------------------
7) Equalities (1) and (2) are absolutely valid for any standard solid, liquid or gaseous conductor.
-----------------------------------------
8) BUT standard liquid and gaseous (and even "vacuum"(!)) conductors have some special features, which are as follows.
----------------------------------------
8A) Liquid conductors. The minimum DC voltage, which is necessary for a standard DC water-splitting electrolysis to occur, is
equal to 1.23 V.
----------------------------------------
8B) Gaseous conductors. The minimum DC voltage, which is necessary for a spark (or arc) to occur, is equal to 2 kV,
where 2 kV = 2,000 V.
---------------------------------------
8C) "Vacuum" conductors. The minimum DC voltage, which is necessary for a direct current to flow through vacuum, is equal to 400 kV, where 400 kV = 400,000 V. This is the case of the so called cold-cathode emission.
---------------------------------------
Note. As if some kind of approximate and (for the present) vague tendency is on its way to be shaped: the lesser the mass dencity and/or hardness of a conductor, the bigger the minimum DC voltage, which is necessary for a direct current to flow through this same conductor.
--------------------------------------
9) A certain portion of hydrogen is generated while standard DC water-splitting electrolysis takes place. And if this portion of hydrogen is burned/exploded, then a certain portion of heat H is generated. Therefore we can write down the equality
H = Z x I x t x HHV (3),
where
H = heat, which is generated if the released hydrogen is burned/exploded
Z = electrochemical equivalent of hydrogen
I = direct current, which flows through the electrolyte while standard DC water-splitting electrolysis takes place
t = time period, within which standard DC water-splitting electrolysis takes place
HHV = higher heating value of hydrogen
---------------------------------------
10) Let us add (3) to the right side of (2) thus forming the inequality
 V x I x t < (I x I x R x t) + (Z x I x t x HHV) (4).
---------------------------------------
11) As a further development of our basic concept let us consider the standard process of recharging of a standard car's battery. In this case in addition to the releasing of hydrogen we store electric energy E. And further, (1) if we disconnect the charger from the already fully charged battery and (2) if we connect the already fully charged battery to a standard copper wire load for example (thus forming a circuit), and (3) if we discharge the battery, then the stored electric energy E transforms entirely into a second additional portion of Joule's heat K. Therefore we can add K to the right side of (4) thus forming the inequality
V x I x t < (I x I x R x t) + (Z x I x t x HHV) + (K) (5).
--------------------------------------
12) In one word, inequalities (4) and (5) unambiguously show that any standard DC water-splitting electrolysis process can be considered as a heater, whose efficiency is bigger than 1.
--------------------------------------
Everything seems to be clear now, doesn't it?
Looking forward to your answers, comments, recommendations, questions.

Free Energy | searching for free energy and discussing free energy

Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #512 on: February 22, 2021, 02:23:40 PM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #513 on: February 22, 2021, 02:30:12 PM »
Symbol 8) in our previuos post to be read as eight (item eight). Some defect of the system transformed eight into an yellow head with black spectacles.

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #514 on: February 23, 2021, 02:43:44 PM »
Any comments, questions, recommendations, related to our last post?

Free Energy | searching for free energy and discussing free energy

Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #514 on: February 23, 2021, 02:43:44 PM »
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Offline Floor

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #515 on: February 24, 2021, 12:53:02 PM »
                                          OHMIIC AND NON OHMIC RESISTANCE

The two most common types of electric resistors are carbon resistors and wire
wound resistors. The amount of current that will flow through these types of
resistors, is in proportion to the voltage applied to them.  They function in accord
with ohms law. (E = I x R, R = E / I and  I = E / R) with the exception that their
electrical resistance increases when their temperature is increased.

E is electromotive force and is measured in units of volts (V).
I is electric current FLOW and is measured in amperes (A).
R is resistance to electric current flow and is measured in units of ohms (Ω).

An electric current flowing through these common types of resistors causes heating
of the resistor.

The heating of these common types of resistors is in proportion to the magnitude of
the current flowing through the resistor.
    But ......
Their resistance increases when their temperature increases.

If the heat generated by the electric current flow, is allowed to dissipate into the
surrounding environment, the resistor may reach an equilibrium between the heat
produced and the heat dissipated.  The magnitude of its electrical resistance is then
stabilized.

Common resistors are rated in terms of the power as heat energy they can dissipate
(their wattage).  Generally the larger the physical size of a resistor, the more heat it
can rapidly dissipate (the more wattage it can handle).  This is because a larger surface
area is in contact with a greater amount of the surrounding air.  Heat transfer is more
rapid when that contact area is greater.

A common electric resistor component AT A GIVEN TEMPERATURE, behaves in accord
with ohm's equations ( E = I x R,  R = E / I  and  I = E / R ). The amount of electric
current that flows through a given resistor, is then in dependent upon the applied
voltage. Its resistance is "ohmic".

Other wise it is "non ohmic", in that Ohm's law does not account for change in its
resistance due to change in temperature.

Example:

A 40 watt, filament type (incandescent) light bulb passes a current of  around 0.33 A)
                     WHEN IT IS AT ITS OPERATING TEMPERATURE.
                       ( P power = E x I), (40 watts / 120 volts equals about 0.33 A)

Its resistance is about  363.6  Ω
                     WHEN IT IS AT ITS OPERATING TEMPERATURE.
                                  (E / I = R), (120 V / 0.33 A = 363 Ω) .

Its resistance when MEASURED at room temperature is around 0.03 Ω.

120 V / 0.03 Ω  =  4000 A. This would result in a (very brief) 480,000 watt power
dissipation (120 V x 4000 A = 480,000 watts), if the the supply system could deliver
that  peak power draw before the filament reaches its operating temperature and
AT WHICH TIME its resistance has increased to 363.6 Ω 

When operated at very low voltage and power input, a 40 watt incandescent lamp
behaves in accord with ohm's equations.  When operated at high voltages and power
input, it is said to behave in a "non ohmic" manner.  The current flowing through it
at a given voltage, has greatly decreased because of its high operating temperature.
Its resistance increased.

Electrolytes behave in accord with ohmic equations at a given temperature except that
there are also chemical reactions which affect and / or affected by temperature and resistance.
                    But also, also .....
the electrical resistance DOES change with simple temperature change.

Electric current through the electrolyte will decrease as the electrolytes resistance
has increased due to increase in its temperature.  But the chemical activity of the
electrolyte will be affected by temperature as well.  Many kinds of chemical reactions
occur faster at higher temperatures than they would occur at lower temperatures.

It seems complex, but to get past a need for temperature stabilization of an electrolyte
during experimentation, one may simply measure power as electric wattage input into
the system, while also accounting for the duration, in time, of the electrolysis.

 Measure...
Power input (wattage). Use a watt meter.

Time duration of the electrolysis. Use a timer.

Temperature of the electrolyte before electrolysis.  Use a thermometer.

Volume of the electrolyte. Use a measuring cup.

Temperature of the electrolyte at the end of the electrolysis Use a thermometer.

Volume of electrolyte remaining after electrolysis.  Use a measuring cup.

Temperature of the evolved gases (including water vapor) at the finish time of the
electrolysis. Use a thermometer.

Volume of the H and O (including water vapor) while under normal atmospheric pressure
and also while at the same temperature as the electrolyte was at, at the end of the
electrolysis.

Remove the water vapor. Pass the gas mixture through a freezing pipe.

How much water was removed from the electrolyte as water vapor. Thaw the ice and
measure its volume with a measuring cup 

Consider this thawed volume of water, to have been heated along with the rest of the
electrolyte and add these calories gained, to those gained from during the rest of
electrolyte's heating during the electrolysis.

Volume of the H and O at regular atmospheric pressure, with out the water vapor, but
also while knowing the temperature of the H and O..

Calculate the calories present within the gas which were gained by the gas during
its temperature increase during the electrolysis.

Calculate the Temperature of the high heat combustion of the H and O produced and
the calories that combustion will produce.

    OR...
look all these things up in a reference source.
« Last Edit: February 24, 2021, 11:18:18 PM by Floor »

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #516 on: February 24, 2021, 02:59:45 PM »
To Floor.
===================
I know that you аre a touchy man and that you get angry very easily. But risking to insult you seriously again I would repeat again: YOU ARE NOT READING MY POSTS AT ALL! AND THIS IS NOT A DIALOGUE AND THIS IS NOT AN HONEST DISCUSSION! THIS IS YOUR MONOLOGUE!
===================
In our post of January 03, 2021, 03:48:09 PM (this post's text has been published many times before in this topic) it is CLEARLY WRITTEN THAT: CONSTANT PURE WATER AND COOLING AGENT SUPPLY COULD KEEP CONSTANT THE ELECTROLYTE'S TEMPERATURE, HEAT EXCHANGE, MASS AND OHMIC RESISTANCE, RESPECTIVELY.
===================
AM I CLEAR NOW? READ CAREFULLY MY POSTS!!!!!



Free Energy | searching for free energy and discussing free energy

Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #516 on: February 24, 2021, 02:59:45 PM »
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Offline Floor

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #517 on: February 24, 2021, 08:39:39 PM »
Yes it is obvious that you do not understand that it is not necessary
to cool the electrolyte, in order that one may conduct the experiment.

In fact cooling the electrolyte would complicate the measurements done
in the experiment.

I understand that you have stated that you do not intend to do experiment.
So your whole topic is without merit anyway, except for the input
from some posters other than your self.

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #518 on: February 26, 2021, 03:47:04 PM »
To Floor.
======================
======================
FIRSTLY.
Prof. S. L. Srivastava's solution is given below.
Prof. S. L. Srivastava's solution consists of two lines only.
LINE 1. Current through the electrolyte is given by I = (m)/(Z x t).
LINE 2. Power consumed = (I) x (I) x (R) =  ((m)/(Z x t)) x ((m)/(Z x t)) x (R) = 31.86 W.
-----------------------------------
I am asking you (PERSONALLY!) my question for the 25th time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!
-----------------------------------
All members of this forum are waiting for your PERSONAL(!) answer for the 24th time. Only one word -- either "yes" or "no"!
======================
======================
SECONDLY.
The three equalities V = I x R (1), V x I x t = I x I x R x t (2) and H = Z x I x t x HHV (3) (that is, Ohm's law, Joule's law of heating and Faraday's laws of electrolysis, respectively) have been validated experimentally millions of times within a period of 200 years. And you reject this simple obvious truth!


Free Energy | searching for free energy and discussing free energy

Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #518 on: February 26, 2021, 03:47:04 PM »
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Offline Floor

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #519 on: February 27, 2021, 03:03:08 PM »
Only on word, yes or no ?
Dream on.

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #520 on: March 01, 2021, 03:58:45 PM »
To Floor.
======================
Prof. S. L. Srivastava's solution is given below.
Prof. S. L. Srivastava's solution consists of two lines only.
LINE 1. Current through the electrolyte is given by I = (m)/(Z x t).
LINE 2. Power consumed = (I) x (I) x (R) =  ((m)/(Z x t)) x ((m)/(Z x t)) x (R) = 31.86 W.
---------------------------------------
I am asking you (PERSONALLY!) my question for the 26th time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!
----------------------------------------
All members of this forum are waiting for your PERSONAL(!) answer for the 26th time. Only one word -- either "yes" or "no"!
======================
CHECKMATE!

Free Energy | searching for free energy and discussing free energy

Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #520 on: March 01, 2021, 03:58:45 PM »
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Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #521 on: March 02, 2021, 03:50:30 PM »
https://endalldisease.com/hyperdrive-free-energy-gerard-morin/
Really amazing link! Where to search for detailed instructions for how to build this device? Any information?

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #522 on: March 04, 2021, 03:44:00 PM »
Any comments related to our standard DC water-splitting electrolysis OU concept?

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #523 on: March 05, 2021, 02:25:15 PM »
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=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
--------------------------
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 "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.
------------------------------

Offline George1

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Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #524 on: March 05, 2021, 02:29:50 PM »
Please read carefully our previous post.
-----------------------------------------
The key to a successful understanding of our OU concept consists of two consequent steps.
-----------------------------------------
Step 1. Prof. S. L. Srivastava's solved problem. (Please read carefully the first part of our previous post.)
-----------------------------------------
Step 2. Our further development of Prof. S. L. Srivastava's solution. (Please read carefully the second part of our previous post.)

 

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