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.
------------------------------
Let me remind again that the text above is a theoretical (only theoretical!) research, which is based however on the most fundamental axioms of electric engineering.
The question is: Do you have any theoretical (only theoretical!) objections against the validity of the theoretical considerations above?
A simple electric heater, which has efficiency greater than 1
Moderator: scott
-
MrTim
- Aficionado
- Posts: 916
- Joined: Thu Nov 06, 2003 11:05 pm
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- Contact:
re: A simple electric heater, which has efficiency greater t
You still leave out the 87% efficiency of electrolysis.
Therefore you are not taken seriously.
That is not theoretical.
Therefore you are not taken seriously.
That is not theoretical.
"....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 Tarsier79 (Nobel prize winner number 1) and to MrTim (Nobel prize winner number 2) .
---------------------------------------
STEP 1.
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 6th 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 6th time. Only one word -- either "yes" or "no"!
---------------------------------------
STEP 1.
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 6th 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 6th time. Only one word -- either "yes" or "no"!
-
MrTim
- Aficionado
- Posts: 916
- 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
In your own words:
You must be using Bill Gates' "There is no wrong math answer" method (You should ask Bill Gates to help with your machine, since the math doesn't matter to him either, because according to him, $2 + $2 <> $4 ;-)
Put the 87% electrolysis efficiency into your equation. Then your equation will be correct...[/i]
And there is the essence of your problem. Your "non-standard manner" is obviously wrong. Your equation bears no resemblance to Srivastava's equation.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.
You must be using Bill Gates' "There is no wrong math answer" method (You should ask Bill Gates to help with your machine, since the math doesn't matter to him either, because according to him, $2 + $2 <> $4 ;-)
Put the 87% electrolysis efficiency into your equation. Then your equation will be correct...[/i]
"....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
Trust me, it still won't be.Put the 87% electrolysis efficiency into your equation. Then your equation will be correct...[/i]
re: A simple electric heater, which has efficiency greater t
To Tarsier79 (Nobel prize winner number 1) and to MrTim (Nobel prize winner number 2).
--------------------------------------
You both are simply two stubborn and ignorant haters. You simply do not understand what are you talking about. There is (a) a ratio released hydrogen/consumed electric energy and (b) a ratio generated Joule's heat/consumed electric energy. Do you make a difference between these two ratios, you ignoramuses? But why am I explaining to you all these things? You both have serious cognitive problems.
---------------------------------------
STEP 1.
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 7th 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 7th time. Only one word -- either "yes" or "no"!
--------------------------------------
You both are simply two stubborn and ignorant haters. You simply do not understand what are you talking about. There is (a) a ratio released hydrogen/consumed electric energy and (b) a ratio generated Joule's heat/consumed electric energy. Do you make a difference between these two ratios, you ignoramuses? But why am I explaining to you all these things? You both have serious cognitive problems.
---------------------------------------
STEP 1.
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 7th 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 7th time. Only one word -- either "yes" or "no"!
re: A simple electric heater, which has efficiency greater t
Seems to be an interesting technology. Please have a look at the link below:
https://endalldisease.com/nitinol-heat- ... ee-energy/
https://endalldisease.com/nitinol-heat- ... ee-energy/
re: A simple electric heater, which has efficiency greater t
Some comments, related to the nitinol heat engine? Seems to be an interesting principle of operation?
re: A simple electric heater, which has efficiency greater t
No interest towards the nitinol heat engine, it's obvious.
In such a case let us get back to our electrolysis.
Any opinions related to the validity of Prof. S. L. Srivastava's solution?
In such a case let us get back to our electrolysis.
Any opinions related to the validity of Prof. S. L. Srivastava's solution?
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.
re: A simple electric heater, which has efficiency greater t
To all members of this forum who are interested in the topic.
-----------------------------
Dear colleagues,
Let me remind again that we (our multinational team) have created 11 (eleven) technology breakthroughs.
-----------------------------
1) Please consider carefully and thoroughly the link below:
https://www.youtube.com/watch?v=aW2ffyvdhjk
-----------------------------
2) The link above describes our first technology breakthrough.
---------------------------
3) The link above describes some simple experiments, which break (a) the law of conservation of mechanical energy and (b) the law of conservation of linear momentum. You can easily carry out these simple experiments in your garage as many times as you want. Any rule/law has its exceptions and there is nothing special, tragic and disturbing in this fact.
---------------------------
4) We (our multinational team) are open to collaboration of mutual benefit (a) for a further perfection and development of our technology breakthroughs and/or (b) for a production of our technology breakthroughs on a large industrial scale.
---------------------------
5) We would like to ask you to popularize the link above as much as possible in internet (and anywhere else and in any possible way).
---------------------------
Let us push forward together the technology progress!
---------------------------
Looking forward to your answer.
Sincerely yours,
PeterAX
-----------------------------
Dear colleagues,
Let me remind again that we (our multinational team) have created 11 (eleven) technology breakthroughs.
-----------------------------
1) Please consider carefully and thoroughly the link below:
https://www.youtube.com/watch?v=aW2ffyvdhjk
-----------------------------
2) The link above describes our first technology breakthrough.
---------------------------
3) The link above describes some simple experiments, which break (a) the law of conservation of mechanical energy and (b) the law of conservation of linear momentum. You can easily carry out these simple experiments in your garage as many times as you want. Any rule/law has its exceptions and there is nothing special, tragic and disturbing in this fact.
---------------------------
4) We (our multinational team) are open to collaboration of mutual benefit (a) for a further perfection and development of our technology breakthroughs and/or (b) for a production of our technology breakthroughs on a large industrial scale.
---------------------------
5) We would like to ask you to popularize the link above as much as possible in internet (and anywhere else and in any possible way).
---------------------------
Let us push forward together the technology progress!
---------------------------
Looking forward to your answer.
Sincerely yours,
PeterAX
-
MrTim
- Aficionado
- Posts: 916
- 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
< c r i c k e t s >
"....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
We (our multinational team) have created 11 (eleven) technology breakthroughs.
-----------------------------
1) Please consider carefully and thoroughly the link below:
https://www.youtube.com/watch?v=xX14NK8GrDY
-----------------------------
2) The link above describes our first technology breakthrough.
---------------------------
3) The link above describes some simple experiments, which break (a) the law of conservation of mechanical energy and (b) the law of conservation of linear momentum. You can easily carry out these simple experiments in your garage as many times as you want. Any rule/law has its exceptions and there is nothing special, tragic and disturbing in this fact.
---------------------------
4) We (our multinational team) are open to collaboration of mutual benefit (a) for a further perfection and development of our technology breakthroughs and/or (b) for a production of our technology breakthroughs on a large industrial scale.
---------------------------
5) We would like to ask you to popularize the link above as much as possible in internet (and anywhere else and in any possible way).
---------------------------
LET US PUSH FORWARD THE TECHNOLOGY PROGRESS!
---------------------------
Looking forward to your answer.
-----------------------------
1) Please consider carefully and thoroughly the link below:
https://www.youtube.com/watch?v=xX14NK8GrDY
-----------------------------
2) The link above describes our first technology breakthrough.
---------------------------
3) The link above describes some simple experiments, which break (a) the law of conservation of mechanical energy and (b) the law of conservation of linear momentum. You can easily carry out these simple experiments in your garage as many times as you want. Any rule/law has its exceptions and there is nothing special, tragic and disturbing in this fact.
---------------------------
4) We (our multinational team) are open to collaboration of mutual benefit (a) for a further perfection and development of our technology breakthroughs and/or (b) for a production of our technology breakthroughs on a large industrial scale.
---------------------------
5) We would like to ask you to popularize the link above as much as possible in internet (and anywhere else and in any possible way).
---------------------------
LET US PUSH FORWARD THE TECHNOLOGY PROGRESS!
---------------------------
Looking forward to your answer.
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.
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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.
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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.
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