Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here: https://overunity.com/5553/privacy-policy/If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

Custom Search

### Author Topic: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1  (Read 252117 times)

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #600 on: April 13, 2021, 05:48:36 PM »
The text below is a slightly modified, shortened and more precise version of our post of March 09, 2021, 02:46:35 PM.
----------------------------
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.)
--------------------------
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.

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #601 on: April 13, 2021, 05:49:29 PM »
EXPERIMENTALLY PROVED reactionless drive and perpetual motion are described in the link below:
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.
----------------------
IMPORTANT NOTE. The technology information described in the link above must be evaluated (SOLELY AND ONLY!) by highly qualified experts (Ph.D.) in theoretical and applied mechanics. Otherwise nothing will come of it.
----------------------

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #602 on: April 14, 2021, 04:41:44 PM »
The link below describes a few simple experiments, which break the law of conservation of mechanical energy and the law of conservation of linear momentum. You can easily carry out these simple experiments in your garage as many times as you want.
---------------------------------------
IMPORTANT NOTE 1. It is highly recommendable that the above mentioned experiments  are evaluated and realized by a highly qualified expert (Ph.D.) in theoretical and applied mechanics.  Otherwise nothing will come of it (most probably).
----------------------------------------
IMPORTANT NOTE 2. The key question in the above mentioned experiments is how to reduce standard friction (where necessary) to a certain minimum limit, beyond which the experimental error (due to friction) is small enough and can be neglected. The answer is simple. You can use for example permanent magnet slides as shown in the link https://www.youtube.com/watch?v=NoW0A8hYs5A . (Permanent magnet slides reduce friction practically to zero and the measuring devices do not register any force of friction.) Aternatively you can use hundreds of other methods for reducing of friction (as much as necessary) as modern technologies allow this feat. We live in 21st century after all.
----------------------------------------
Looking forward to your comments after repeating the above mentioned simple experiments.

#### lancaIV

• elite_member
• Hero Member
• Posts: 5233
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #603 on: April 15, 2021, 03:00:40 PM »
George1, where begins and where ends "simple heater" ?

2008 : http://www.borderlands.de/net_pdf/NET1108S39-41.pdf  Wärmegenerator von Hans Peter Bierbaumer

2013 : http://www.borderlands.de/net_pdf/NET0313S10-18.pdf  E3 Gravitherme mit COP = 50:1 - ein Zukunftsprojekt

translated "device"-claim :

" ......  In the prospectus description for E3 Gravitherme says that this heat generation system with
a COP> 50 is unique in the world.
A device with a heating output of 20 kW therefore requires less than 400 W connected load and
can easily be converted into a existing flow / return linea heater can be installed.

A Exhaust gas chimney is not required.
On Gas or oil can then with one of those Heating practically omitted entirely and the operating costs are exceptionally low. .... "

Consumer of elec. energy as per voltmeter and  ampere meter displays E1 = I × V × Δτ ,KJ    0.86  0.86  2.70  2.43

Energy of the heated solution,                                                            E2 = 4, 19 × m × Δt, kJ  27.53  30.72  87.05  48.43

Efficiency level of the cell as per voltmeter and  ampere meter displays K = E2/E1     32.01  35.70  32.24  33.32

let us go to
https://worldwide.espacenet.com/publicationDetails/biblio?FT=D&date=20120613&DB=EPODOC&locale=en_EP&CC=EP&NR=1875140B1&KC=B1&ND=4

there above written EP Register,entering :

https://register.epo.org/application?lng=en&number=EP05731926&tab=main

https://data.epo.org/publication-server/pdf-document?pn=1875140&ki=B1&cc=EP&pd=20120613

Veröffentlichungstag und Bekanntmachung des Hinweises auf die Patenterteilung: 13.06.2012 Patentblatt 2012/24

Publication date and announcement of the reference to the  patent granted: 13.06.2012  Patentblatt 2012/24

technical data :

electric input :   Verbraucher an el. Energie laut Voltmeter- und Ampermeteranzeigen E1=IxVx Δτ ,kJ  0,86  0,86  2,70  2,43

thermic output :Energie der erhitzten Lösung, E2= 4,19 xmx Δt, kJ  27,53  30,72  87,05  48,43

efficiency :      Wirkungsgrad der Zelle laut Voltmeter- und Ampermeteranzeigen  K=E2/E1  32,01  35,70  32,24  33,32

=                                                      between 3201% and 3570% conversion

Patentoffice peers acceptable numbers : granted !

But http://www.borderlands.de/net_pdf/NET0313S10-18.pdf ,ATTENTION,page 18 :

The “Austrian Society for critical thinking ”, by the way the inventor of the Gravitherme for the Award “The Golden Board vorm Kopf ”nominated for consistently ignoring the
Thermodynamics in energy multiplication and the pseudoscientific painting of his patent application

Whom we can trust ? We have to think and decide : critically ! After device teste and measurement !

Btw : https://register.epo.org/application?number=EP05731926&lng=en&tab=event
11.05.2012                   (Expected) grant           published on  13.06.2012  [2012/24]

https://www.wienerzeitung.at/nachrichten/wirtschaft/oesterreich/494114_Wasserstoff-Technologie-Firma-Hydrogen-Research-AG-pleite.html

Hydrogen technology company Hydrogen Research AG bankrupt

insolvency

After the death of founder Hans-Peter Bierbaumer, insolvent AG is wound up.

https://totenbilder.at/totenbild/bierbaumer-hans-peter  R.I.P. since 19.08.2012

granted after the applicant his death !

But: see EP Patent document !

Inventor(s):MICHAYLOVICH KANAREV PHILIPP [RU]

and we see that the bureaucracy works/(re-)act really slow :

article date from October 15, 2012, 3:45 pm Clock

related

Event date :2013/10/31

Event code :BERE

Code Expl.:-   BE: LAPSED

NEW OWNER :BIERBAUMER, HANS-PETER DR. H.C.
EFFECTIVE DATE :20130430 (    more than 1 year after his death  , "honoris causa"  )

2008,....,2013,.....2020,2021,.....

E3 Gravitherme mit COP = 50:1 - ein Zukunftsprojekt             E3 Gravitherme with COP = 50: 1 - a future project

« Last Edit: April 15, 2021, 07:11:54 PM by lancaIV »

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #604 on: April 18, 2021, 03:27:29 PM »
To lancaIV.
==================================
Ok, let us focus on the electric heater again.
==================================
The text below is a slightly modified, shortened and more precise version of our post of March 09, 2021, 02:46:35 PM.
----------------------------
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.)
--------------------------
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.
-----------------------------

#### lancaIV

• elite_member
• Hero Member
• Posts: 5233
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #605 on: April 18, 2021, 08:45:27 PM »
a. it is not a real life experiment

b. it is a tutorial question/answer statement ,with idealistic = theoretical ideal,lost-free,numbers

c. it is not hydrolysis,it is not thermolysis : it is " In the electrolysis of sulphuric acid solution" related

Known best pure electric input +

f.e.
https://www.chemistryworld.com/news/magnets-that-double-efficiency-of-water-splitting-could-help-usher-in-a-hydrogen-economy/3010618.article

Not new experimental knowledge :

#### NdaClouDzzz

• Sr. Member
• Posts: 305
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #606 on: April 23, 2021, 12:35:12 AM »
The free electrons in the electrical circuit which produced the electron current is finite. That is, by Faradays law all the electrons which left the (-) negative terminal must complete the circuit and return to the (+) positive terminal of the source. This is also the premise of what is known as Kirchhoff's laws saying we cannot gain or lose anything in the closed circuit. No electrons can be lost or gained to the electrolyte and energy/mass are conserved. However another law also applies which states the electron current must dissipate all it's energy as heat in the load resistance or the source resistance for energy to be conserved.
Thus we seem to have a contradiction where one law claims X must happen and another law which claims Y must happen but X-Y contradict one another.

Nice catch, indeed!
Have you resolved the contradiction? If so, please explain how, as to me it is a head-scratcher!

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #607 on: April 24, 2021, 12:50:36 PM »
To lancaIV.
==================================
Ok, let us focus on the electric heater again.
==================================
The text below is a slightly modified, shortened and more precise version of our post of March 09, 2021, 02:46:35 PM.
----------------------------
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.)
--------------------------
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.

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #608 on: April 24, 2021, 12:54:47 PM »
To lancaIV.
======================
Prof. S. L. Srivastava's solution is given in our previous post.
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 1st 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 1st time. Only one word -- either "yes" or "no"!

#### lancaIV

• elite_member
• Hero Member
• Posts: 5233
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #609 on: April 24, 2021, 01:00:17 PM »

You tried this funny "game" with Floor,now trial 2.0 ! Your chance : up to trial  84053,0 following actual number :  Members: 84053

Happy weekend

OCWL

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #610 on: April 24, 2021, 01:13:58 PM »
To lancaIV.
===================
You totally disappont me. You copy entirely the stupid and unskillful Floor's manipulation methods. And it is normally my counter-approach to be the same. Shame on you!

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #611 on: April 24, 2021, 01:14:53 PM »
To lancaIV.
======================
Prof. S. L. Srivastava's solution is given in our today's post of 12:54:47 PM.
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 2nd 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 2nd time. Only one word -- either "yes" or "no"!

#### lancaIV

• elite_member
• Hero Member
• Posts: 5233
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #612 on: April 24, 2021, 01:27:19 PM »
Sie verlieren so viel persoenliche Zeit,George1 !

Shame on you!    Scham !    Welch Schwaeche ! Okay, eine Volks-Trauer-Scham-Nano-Sekunde  : Start-Stop ! Genug geschaemt !

Vielleicht sollte ein " Shame on You/ deontologisch:Fegefeuer/chama "- button installiert werden

Ich vermute und ziehe deshalb schon vor :

I am asking you (PERSONALLY!) my question for the 3rd time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!

und weil es dann so schoen lang werden koennte :

I am asking you (PERSONALLY!) my question for the ......... time: Is Prof. S. L. Srivastava's solution correct? Yes or no? Only one word -- either "yes" or "no"!

Nach 3 kommt 4 , wenn nicht auf Kommata zurueckgegriffen werden sollte : 3,1 ; 3,11 ; 3,111; ......

All members of this forum are waiting for your PERSONAL(!) answer for the 3rd time. Only one word -- either "yes" or "no"!

"All members ..... " : ich gebe hiermit bekannt,dass ich,als Teil des "All", darauf verzichte zu warten ! BASTA ,nochmoi !

Und ein "persoenliches : Ja ! " beziehungsweise "persoenliches : Nein !"  :

ICH SO WIE SIE WERDEN NIEMALS INNERHALB EINES DIGITALEN FORUMS PERSOENLICH WERDEN KOENNEN
( so unter virtuellen 4 Ohren gesagt )

"Only one word" :               oh,da fuehlt man sich Dekaden zurueckgeworfen,Katechese betreffend  :

Herr(schaften),        nur ein Wort ,                 so wird UNSER(e) Seel(e) gesund

ALS MENSCH VERFUEGT MAN UEBER VIEL ZEIT, DABEI FREI VON ULTIMATEN SOWIE MORATORIEN

Ein schoenes Wochenende wuenschend

OCWL

p.s.: you are refering a Prof. S. L. Srivastava tutorial (hand book) ,based by (rhetorical/specific moment) question/answer examples

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #613 on: April 26, 2021, 12:13:29 PM »
To lancaIV.
======================
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 3rd 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 3rd time. Only one word -- either "yes" or "no"!

#### George1

• Hero Member
• Posts: 884
##### Re: A SIMPLE ELECTRIC HEATER, WHICH HAS EFFICIENCY GREATER THAN 1
« Reply #614 on: April 29, 2021, 09:18:37 AM »
The text below is a copy of our post of April 24, 12:50:36 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.)
--------------------------
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.
-----------------------------