PeterAX 0 Posted March 10 Report Share Posted March 10 Dear colleagues, Please look 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=rr...ectrochemical equivalent of hydrogen"&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. ------------------------------ 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 COP = 1.37, that is, we have again COP > 1. =================== The text above can be also found in the three links below: https://www.physicsforums.com/threads/a-not-related-to-the-heat-pump-process-of-cop-1.1000688/ https://overunity.com/18134/a-simpl...-has-efficiency-greater-than-1/msg555939/#new https://besslerwheel.com/forum/viewtopic.php?t=8017&postdays=0&postorder=asc&start=510 =================== What do you think about the concept above? What is your opinion? Looking forward to your answer. Quote Link to post Share on other sites

PeterAX 0 Posted March 12 Author Report Share Posted March 12 The text below is a slightly modified, shortened and more precise version of our previous post of Wednesday at 05:13 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=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. 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. ------------------------------ The text above can be also found in the three links below: https://www.physicsforums.com/threads/a-not-related-to-the-heat-pump-process-of-cop-1.1000688/ https://overunity.com/18134/a-simple-electric-heater-which-has-efficiency-greater-than-1/msg555985/#new https://besslerwheel.com/forum/viewtopic.php?t=8017&postdays=0&postorder=asc&start=525 =================== What do you think about the concept above? What is your opinion? Looking forward to your answer. ======================= P.S. 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? Quote Link to post Share on other sites

PeterAX 0 Posted March 19 Author Report Share Posted March 19 Any comments, questions, recommendations? Quote Link to post Share on other sites

PeterAX 0 Posted March 27 Author Report Share Posted March 27 We keep preparing the YouTube clip, which illustrates our basic concepts. The clip will contain description of REAL EXPERIMENTS, which you can carry out in your garage as many times as you want. Quote Link to post Share on other sites

PeterAX 0 Posted April 5 Author Report Share Posted April 5 Dear colleagues, 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 together the technology progress! --------------------------- Looking forward to your answer. Sincerely yours, Peter Axe Quote Link to post Share on other sites

PeterAX 0 Posted April 6 Author Report Share Posted April 6 The text below is a slightly modified, shortened and more precise version of our previous post of Wednesday at 05:13 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=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. 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. Quote Link to post Share on other sites

PeterAX 0 Posted April 6 Author Report Share Posted April 6 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! Quote Link to post Share on other sites

PeterAX 0 Posted April 8 Author Report Share Posted April 8 EXPERIMENTALLY PROVED reactionless drive and perpetual motion are described in the link below: https://www.youtube.com/watch?v=xX14NK8GrDY&ab_channel=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. Quote Link to post Share on other sites

Dubbelosix 152 Posted April 9 Report Share Posted April 9 Ocean please get this where it belongs. OceanBreeze 1 Quote Link to post Share on other sites

PeterAX 0 Posted April 10 Author Report Share Posted April 10 To Dubbelosix. ------------------------------- Sit down and read many, many times some beginners' course/guide of elementary physics! Because otherwise you resemble a clown! Quote Link to post Share on other sites

Dubbelosix 152 Posted April 10 Report Share Posted April 10 The calculations look right..... Quote Link to post Share on other sites

Dubbelosix 152 Posted April 10 Report Share Posted April 10 (edited) After some thorough looking however it's not all too clear. Some of this may still be wrong. Edited April 10 by Dubbelosix Quote Link to post Share on other sites

Dubbelosix 152 Posted April 10 Report Share Posted April 10 (edited) Neither am I saying it is wrong but instead it should be provably wrong in theory. Edited April 10 by Dubbelosix Quote Link to post Share on other sites

OceanBreeze 425 Posted April 10 Report Share Posted April 10 This is very similar to a heat pump. You are inputting 38232 J of energy in order to extract a total of 51646 J of energy, which includes 14200 J of energy provided by the burning of hydrogen, an external input! This is exactly the way a classical heat pump works! In fact, a COP greater than one is the rule, rather than the exception and your COP of 1.35 is very low! A classical heat pump routinely achieves a COP of around 4.5 This is more nonsense and I will move it to where it belongs, in Silly Claims. This is your last warning for posting nonsense and annoying other members. Dubbelosix 1 Quote Link to post Share on other sites

Dubbelosix 152 Posted April 10 Report Share Posted April 10 Right this is better than I should have articulated. Quote Link to post Share on other sites

PeterAX 0 Posted April 18 Author Report Share Posted April 18 (edited) To OceanBreeze. ------------------------------ The standard classic heat pump has nothing to do with our new electric heater. The two devices have two entirely different principles of operation. Please read carefully and thoroughly our posts, if possible. Looking forward to your answer. Edited April 18 by PeterAX Quote Link to post Share on other sites

PeterAX 0 Posted April 26 Author Report Share Posted April 26 Please look again at our post of 04/06/2021 03:59 PM. --------------------------------------------------------- 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 only a simple question: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"! ---------------------------------------- Waiting for your simple answer. Only one word -- either "yes" or "no"! Quote Link to post Share on other sites

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