https://www.youtube.com/watch?v=xX14NK8GrDY
----------------------------
The link above clearly shows how a few simple experiments, carried out in your garage, (1) can be the basis for designing of a simple mechanical reactionless drive machine and (2) can solve your personal energy problems (as well as the energy problems of the world as whole).
-----------------------------
Let us popularize the link above as much as possible in internet (and anywhere else and in any other way).
LET US PUSH FORWARD TOGETHER THE TECHNOLOGY PROGRESS!
A simple electric heater, which has efficiency greater than 1
Moderator: scott
re: A simple electric heater, which has efficiency greater t
EXPERIMENTALLY PROVED reactionless drive and perpetual motion are described in the link below:
https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe
The link above describes a few simple reactionless drive and perpetual motion experiments. You can easily carry out these simple experiments in your garage as many times as you want.
Looking forward to your opinions, recommendations, questions.
https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe
The link above describes a few simple reactionless drive and perpetual motion experiments. You can easily carry out these simple experiments in your garage as many times as you want.
Looking forward to your opinions, recommendations, questions.
re: A simple electric heater, which has efficiency greater t
Still no comments related to the link below?
https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe
--------------------------
What happens here? Where is the new technologies pioneer spirit here in this forum?:)
--------------------------
Looking forward to your answers/comments.
https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe
--------------------------
What happens here? Where is the new technologies pioneer spirit here in this forum?:)
--------------------------
Looking forward to your answers/comments.
re: A simple electric heater, which has efficiency greater t
The text below is a slightly modified, shortened and more precise version of our post of Fri Mar 05, 2021 12:47 pm.
----------------------------
Please 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.
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.
---------------------------
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) The Joule's heat, generated in the process of electrolysis is given by
Q = (I) x (I) x (R) x (t) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) x (t) = (31.86 W) x (1200 s) = 38232 Ws = 38232 J = outlet energy 1.
3) 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 = (HHV) x (m) = (142 000 000) x (0.0001) = 14200 J = outlet energy 2,
where
m = mass of the released hydrogen
HHV = higher heating value oh hydrogen
4) Therefore we can write down the equalities:
4A) outlet energy 1 + outlet energy 2 = 38232 J + 14200 J = 52432 J
4B) inlet energy = 38232 J.
5) Therefore COP is given by
COP = 52432 J/38232 J = 1.37 <=> COP = 1.37 <=> COP > 1.
------------------------------
IMPORTANT NOTE. 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 the same COP = 1.37, that is, we have again the same COP > 1.
----------------------------
Please 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.
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.
---------------------------
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) The Joule's heat, generated in the process of electrolysis is given by
Q = (I) x (I) x (R) x (t) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) x (t) = (31.86 W) x (1200 s) = 38232 Ws = 38232 J = outlet energy 1.
3) 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 = (HHV) x (m) = (142 000 000) x (0.0001) = 14200 J = outlet energy 2,
where
m = mass of the released hydrogen
HHV = higher heating value oh hydrogen
4) Therefore we can write down the equalities:
4A) outlet energy 1 + outlet energy 2 = 38232 J + 14200 J = 52432 J
4B) inlet energy = 38232 J.
5) Therefore COP is given by
COP = 52432 J/38232 J = 1.37 <=> COP = 1.37 <=> COP > 1.
------------------------------
IMPORTANT NOTE. 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 the same COP = 1.37, that is, we have again the same COP > 1.
-
MrTim
- Aficionado
- Posts: 925
- Joined: Thu Nov 06, 2003 11:05 pm
- Location: "Excellent!" Besslerwheel.com's C. Montgomery Burns
- Contact:
re: A simple electric heater, which has efficiency greater t
ZZZZZzzzzzzzzzzzzzzzzzzz.....
"....the mechanism is so simple that even a wheel may be too small to contain it...."
"Sometimes the harder you look the better it hides." - Dilbert's garbageman
-
Georg Künstler
- Devotee
- Posts: 1718
- Joined: Fri Nov 07, 2003 12:22 pm
- Location: Speyer, Germany
- Contact:
re: A simple electric heater, which has efficiency greater t
Hi PeterAX,
energy can only be converted from one form to another.
So look for the gap in your calculation and formula.
An efficiency greater than 1 requires 2 energy sources.
one energy source and a pumping energy source so that you can combine the two energy source for your advantage.
But it is a conversion, transform of energy, you will never get a gain.
Think of a heat pump, for example.
What we try here on the Besslerwheel.com board is to get Gravity work for us so that we have an advantage.
That is different to your task.
Gravity is around us, so we can use Gravity force on every place in the universe.
Can you help us to convert the gravitational force into useful energy
energy can only be converted from one form to another.
So look for the gap in your calculation and formula.
An efficiency greater than 1 requires 2 energy sources.
one energy source and a pumping energy source so that you can combine the two energy source for your advantage.
But it is a conversion, transform of energy, you will never get a gain.
Think of a heat pump, for example.
What we try here on the Besslerwheel.com board is to get Gravity work for us so that we have an advantage.
That is different to your task.
Gravity is around us, so we can use Gravity force on every place in the universe.
Can you help us to convert the gravitational force into useful energy
Best regards
Georg
Georg
re: A simple electric heater, which has efficiency greater t
To Georg Kunsler.
----------------------------------
Hi there,
Thank you for your reply.
1) You wrote: "That is different to your task. Gravity is around us, so we can use Gravity force on every place in the universe. Can you help us to convert the gravitational force into useful energy?"
2) The answer is simple. Please study carefully again and step by step, and page by page the link I have sent in my previous posts. Carry out the simple experiments in your garage and simply measure heights, masses and velocities and from here calculate potential and kinetic energies. It's simple.
(Here is the link again: https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe )
----------------------------------
Looking forward to your comments after repeating the few simple experiments described in the link above.
Regards,
----------------------------------
Hi there,
Thank you for your reply.
1) You wrote: "That is different to your task. Gravity is around us, so we can use Gravity force on every place in the universe. Can you help us to convert the gravitational force into useful energy?"
2) The answer is simple. Please study carefully again and step by step, and page by page the link I have sent in my previous posts. Carry out the simple experiments in your garage and simply measure heights, masses and velocities and from here calculate potential and kinetic energies. It's simple.
(Here is the link again: https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe )
----------------------------------
Looking forward to your comments after repeating the few simple experiments described in the link above.
Regards,
-
MrTim
- Aficionado
- Posts: 925
- Joined: Thu Nov 06, 2003 11:05 pm
- Location: "Excellent!" Besslerwheel.com's C. Montgomery Burns
- Contact:
Re: re: A simple electric heater, which has efficiency great
...
"....the mechanism is so simple that even a wheel may be too small to contain it...."
"Sometimes the harder you look the better it hides." - Dilbert's garbageman
re: A simple electric heater, which has efficiency greater t
To Georg Kunstler.
-------------------------------
Hi Georg,
Still no answer from you? What happens?
It seems to me that the Bessler wheel and the effect, described in the link https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe , can be successfully combined and work together. What do you think about this?
Looking forward to your answer.
Regards,
-------------------------------
Hi Georg,
Still no answer from you? What happens?
It seems to me that the Bessler wheel and the effect, described in the link https://www.youtube.com/watch?v=xX14NK8 ... l=PeterAxe , can be successfully combined and work together. What do you think about this?
Looking forward to your answer.
Regards,
re: A simple electric heater, which has efficiency greater t
The text below is a copy of our post of April 13, 2021, 5:32 pm. (The text below has been published many times in this forum topic.)
----------------------------
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.
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.
---------------------------
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) The Joule's heat, generated in the process of electrolysis is given by
Q = (I) x (I) x (R) x (t) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) x (t) = (31.86 W) x (1200 s) = 38232 Ws = 38232 J = outlet energy 1.
3) 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 = (HHV) x (m) = (142 000 000) x (0.0001) = 14200 J = outlet energy 2,
where
m = mass of the released hydrogen
HHV = higher heating value oh hydrogen
4) Therefore we can write down the equalities:
4A) outlet energy 1 + outlet energy 2 = 38232 J + 14200 J = 52432 J
4B) inlet energy = 38232 J.
5) Therefore COP is given by
COP = 52432 J/38232 J = 1.37 <=> COP = 1.37 <=> COP > 1.
------------------------------
IMPORTANT NOTE. 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 the same COP = 1.37, that is, we have again the same COP > 1.
-----------------------------
IMPORTANT NOTE. The text above must be evaluated (SOLELY AND ONLY!) by highly qualified experts (Ph.D.) in electric engineering. Otherwise nothing will come out of it.
-----------------------------
Looking forward to your comments.
----------------------------
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.
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.
---------------------------
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) The Joule's heat, generated in the process of electrolysis is given by
Q = (I) x (I) x (R) x (t) = ((m)/(Z x t)) x ((m)/(Z x t)) x (R) x (t) = (31.86 W) x (1200 s) = 38232 Ws = 38232 J = outlet energy 1.
3) 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 = (HHV) x (m) = (142 000 000) x (0.0001) = 14200 J = outlet energy 2,
where
m = mass of the released hydrogen
HHV = higher heating value oh hydrogen
4) Therefore we can write down the equalities:
4A) outlet energy 1 + outlet energy 2 = 38232 J + 14200 J = 52432 J
4B) inlet energy = 38232 J.
5) Therefore COP is given by
COP = 52432 J/38232 J = 1.37 <=> COP = 1.37 <=> COP > 1.
------------------------------
IMPORTANT NOTE. 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 the same COP = 1.37, that is, we have again the same COP > 1.
-----------------------------
IMPORTANT NOTE. The text above must be evaluated (SOLELY AND ONLY!) by highly qualified experts (Ph.D.) in electric engineering. Otherwise nothing will come out of it.
-----------------------------
Looking forward to your comments.
-
MrTim
- Aficionado
- Posts: 925
- Joined: Thu Nov 06, 2003 11:05 pm
- Location: "Excellent!" Besslerwheel.com's C. Montgomery Burns
- Contact:
re: A simple electric heater, which has efficiency greater t
PeterAX writes:
The text below has been published many times in this forum topic.
"....the mechanism is so simple that even a wheel may be too small to contain it...."
"Sometimes the harder you look the better it hides." - Dilbert's garbageman
re: A simple electric heater, which has efficiency greater t
To MrTim.
-----------------------------
Your stupid animations cannot hide the truth. I really wonder why do you take part in the discussions of this forum. Because it is more than evident that you are an absolute amateur in any branch of physics.
=======================
=======================
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 8th time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!
----------------------------------------
I am waiting for your PERSONAL(!) answer for the 8th time. Only one word -- either "yes" or "no"!
-----------------------------
Your stupid animations cannot hide the truth. I really wonder why do you take part in the discussions of this forum. Because it is more than evident that you are an absolute amateur in any branch of physics.
=======================
=======================
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 8th time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!
----------------------------------------
I am waiting for your PERSONAL(!) answer for the 8th time. Only one word -- either "yes" or "no"!
re: A simple electric heater, which has efficiency greater t
Where is the brave new technologies pioneer spirit in this forum? What happens here? :)
re: A simple electric heater, which has efficiency greater t
You simply have to accept the truth.
re: A simple electric heater, which has efficiency greater t
Still no comments? As if nobody in this forum wants to save some money while heating his/her house during winter season?:)