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Author Topic: quentron.com  (Read 1261588 times)

MarkE

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Re: quentron.com
« Reply #1980 on: February 03, 2014, 01:48:09 AM »

Sorry MarkE but your comments that you believe academics would run to ring the bell of discovery are wrong, I know for I have had the conversations many times, you are merely speculating. The history of science is that people are attacked for holding a contrary view, in fact almost all discoveries are ignored for some time until some critical mass is achieved. That is the way it is and your comment about it ringing wrong is part of the problem, you assume I am wrong because others are not on the front page of the news.

You are not going to do yourself any good by either misquoting those you converse with, or by asserting what you think is in their minds.  My experience has been that those who have come upon significant discoveries and who have verified those discoveries have been very anxious to report them.  What I use to assess the correctness of an idea is the direct evidence that supports or refutes the idea.  So far you seem to express the idea that different work functions can be manipulated to create an energy source.  That idea runs afoul of established evidence concerning the behavior of work functions.  I will read the document you linked and see what it has to say.
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 You should try for yourself to get a science journal to return your call if you send them a note saying you have just found a way of converting ambient heat to power with 100% efficiency, or that you have breached the Kelvin interpretation, if they reply they will tell you you must have made a mistake.
What do you mean when you say: "converting ambient heat to power with 100% efficiency"? 

Do you think you can convert an isothermal reservoir into two reservoirs:  one at a higher temperature and one at a colder temperature than the original without an external energy source?  Such a feat would be quite remarkable and demand very strong evidence.  Certainly, the extraordinary nature of the claim strongly suggests that error is probable.
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You and others on these sites also tell me I must have made a mistake.

Again, I have as of this time said no such thing, and I will thank you to not misquote me.  I do ask that you provide evidence commensurate to your extraordinary claim.
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The next bit where you give me a lecture is downright rude, the fact is I know this science clearly better than you do,

Mr. Hardcastle you have so far both misquoted me and told me what is in my mind, and unilaterally declared some personal superiority.  Kindly restrict yourself to technical discussion.
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I never said anywhere that work function is an energy source, I fully understand all the science. I will not take you to task with what you said or gesture to educate you for I have better things than to engage in a Profitis / Sarkeizen style exchange, and I am sure you do to.

MileHigh

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Re: quentron.com
« Reply #1981 on: February 03, 2014, 01:53:23 AM »
Philip:

Somebody around here that is long gone ran that experiment in a toaster oven!  A bloody toaster oven and you endorsed their results!  That's not an isothermal environment at all.  In fact it was more of a nightmare for the experimenter because the toaster oven itself started smoking and started to melt down.

The fact that you would endorse that farcical "experiment" does not give me the warm and fuzzies.  I also seem to recall that from about two years ago the claim was something like 100 or perhaps 1000 amperes per square centimeter but you never stated the voltage output.  So for me the alleged current density claim and the ignoring of translating that into a power output claim is not confidence inspiring.  We are assuming that you have a very good heat reservoir, like a high rate of water passing across your device (like a water cooling system for a high-end PC).  But that was never discussed by you either as far as I can recall.

Another issue that never really got seriously discussed is that if you actually had working semiconductor devices, then there is only so much thermal power that can flow into the device based on the thermal resistance of the ambient medium that the chip finds itself in.  We can assume an infinite heat reservoir.  So assuming 100% efficiency, them the rate of thermal power that lands on the device and gets converted into electricity is a major concern.  This would choke the available power output from your alleged device also.  You can imagine "cold spots," thermal energy sinks, due to the existence of your device.  Effectively it could become a thermal-electrical perpetual motion machine:  "thermal sink -> electrical -> resistive heater -> thermal sink."  Some people posted taking it very seriously saying that a cell phone could power itself by putting one of your chips on top of the main processor chip.

I realize that I am mostly discussing the "mechanics" here and they are secondary to proving the device works as claimed.  However, there were many wild claims made about your device assuming that it actually worked that totally ignored these "thermal mechanics" issues. I note that you never seemed to be interested in addressing these issues yourself.

Free energy cars powered by ambient heat that leave a wake of extremely cold air as they drive down the highway?  I don't think so.  (Note the car would also leave a wake of heated air that would be in balance with the extremely cold air.)

Meanwhile we wait for the timing of the "next" goal post.

MileHigh

MarkE

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Re: quentron.com
« Reply #1982 on: February 03, 2014, 03:11:43 AM »
Here is the one pager I posted somewhere else.
Mr. Hardcastle please tell me that you have much more than that one page pdf file.  That document asserts a claim that the experiment: "violates Kelvin's interpretation of the Second Law of Thermodynamics".  It fails to even quote the statement it purports to refute, much less attempt to justify how it would manage to do that.  The Kelvin statement of the second law refers to a single thermal reservoir.  In the experiment as it is sparsely outlined, there are two heat reservoirs in the system and whatever heat source that drives the oven. 

Philip Hardcastle

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Re: quentron.com
« Reply #1983 on: February 03, 2014, 03:49:21 AM »
Mr. Hardcastle ...... In the experiment as it is sparsely outlined, there are two heat reservoirs in the system....




You have to be joking?


The fact that the measuring moving coil meter is at room temperature is completely immaterial. The DUT is immersed in a single heat reservoir, it has no temperature gradient across it. Both of the wires connecting the DUT to the outside world are of identical metal, the meter terminals are at the same temperature. What power drives the oven in also immaterial.


I am not going to waste any more time chatting with you.




MarkE

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Re: quentron.com
« Reply #1984 on: February 03, 2014, 04:54:43 AM »



You have to be joking?


The fact that the measuring moving coil meter is at room temperature is completely immaterial. The DUT is immersed in a single heat reservoir, it has no temperature gradient across it. Both of the wires connecting the DUT to the outside world are of identical metal, the meter terminals are at the same temperature. What power drives the oven in also immaterial.


I am not going to waste any more time chatting with you.
Mr. Hardcastle, in order to violate Lord Kelvin's expression of the Second Law of Energy, you need to do work by removing heat from a single reservoir.  More specifically, you must perform work with every calorie that you remove from that single reservoir.

The Kelvin statement of the Second Law of Energy is:
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"There is no process whose only effect is to accept heat from a single reservoir and transform it entirely into work."

Your apparatus has much more than just a single heat reservoir.  It has at least two reservoirs and an input power source.

Philip Hardcastle

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Re: quentron.com
« Reply #1985 on: February 03, 2014, 05:22:26 AM »
Mr. Hardcastle, in order to violate Lord Kelvin's expression of the Second Law of Energy, you need to do work by removing heat from a single reservoir.  More specifically, you must perform work with every calorie that you remove from that single reservoir.

The Kelvin statement of the Second Law of Energy is:
Your apparatus has much more than just a single heat reservoir.  It has at least two reservoirs and an input power source.


It, the device under test, does not operate between two reservoirs, it does not operate with a temperature difference across its thickness, end of argument.


It, the single reservoir, receives ac input power only because we lose heat through the oven walls to ambiebt, if the DUT was in a natural 550C reservoir we would not be having this stupid conversation about input power to an oven.


MarkE, if you insist on saying the DUT is in two reservoirs then I have to believe you are being deliberately obtuse.


I will not respond to further silly statements on this thread, if you have something sensible to say you can email me.


The plot was done by a Canadian physicist from the university of Montreal, he carefully measured in upward and downward paused steps, the curves are identical in both directions.


Result have been obtained in a few other labs with near identical results.


There have been 2 experiments where the voltage was produced but with little current, these are simply the result of using old tubes with cathode work function higher than the spec due to them being 30 year old devices.


BTW if you look at the construction of a pentode the cathode is completely surrounded by the anode, so when the tube is being heated it is 100% certain that the cathode is not hotter than the anode.


All my experiments showed summed currents with DUT's in parallel, added voltages when done in series, and all done with extreme temperature control. Also done was the DUT shorted out which gives zero outputs.


Apart from the Chief Scientist of a major technology company there was also a gentleman here that modified his toaster oven which managed to get the temperature to 500C plus, he did apparently get smoking wires that made him ill, and he did get an output the same as many others, was it a proof? no of course not, but it was a genuine attempt to try to replicate rather than to do nothing, which is what all skeptics do, absolutely nothing constructive.


I again suggest you use the pentode, once you understand the single reservoir and thermocouple theory issues, and use your access to equipment and an hour of your time to do the simple experiment, then when you get the 850mV output and see a current of 5uA you can go about showing us how you can account for it with the status quo.




sarkeizen

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Re: quentron.com
« Reply #1986 on: February 03, 2014, 05:25:41 AM »
zero incompatibility means we can use a college textbook to predict this repeatable cycle.
Yawn.

Well, a couple of things:

i) You haven't presented any formal argument demonstrating that there's zero incompatibility.  You just assumed it.  Assuming something isn't the same as proving it.  This should be obvious to even a stupid person.

ii) Even if you could demonstrate something like there is "no contradiction" between the two position.   The fact that there exists no contradiction between two principles (A & B) i.e. A does not imply !B and B does not imply !A.  That is actually NOT the same as saying A implies B.  Again something you would know if you had more than a 5th grade education.

sarkeizen

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Re: quentron.com
« Reply #1987 on: February 03, 2014, 05:48:10 AM »
was it a proof? no of course not, but it was a genuine attempt to try to replicate rather than to do nothing, which is what all skeptics do, absolutely nothing constructive.
So much double-speak so little time.  I truly don't understand this kind of talk at all.  So "doing nothing" is bad.  However attempting to replicate an effect in an environment so error prone that you can't really tell if he did it or not is "doing something".

I really have to ask..."In what sense".  While he was doing that I was probably playing X-Box.  While we could both be considered "doing something" in a broad sense.  I'm not sure either of us could be considered "usefully attempting to replicate the alleged sebby effect".

MarkE

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Re: quentron.com
« Reply #1988 on: February 03, 2014, 07:02:31 AM »
Mr. Hardcastle, there are thermal leaks all over your experiment.  That seems to make it impossible for your experiment to test whether heat taken from a reservoir can be completely converted to work contrary to the Kelvin statement.

Philip Hardcastle

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Re: quentron.com
« Reply #1989 on: February 03, 2014, 08:36:53 AM »
Mr. Hardcastle, there are thermal leaks all over your experiment.  That seems to make it impossible for your experiment to test whether heat taken from a reservoir can be completely converted to work contrary to the Kelvin statement.


I am at a loss as to how you think, but anyway.


Of course it is going to be very difficult to show heat taken in by the DUT to electrical output unless we build around the DUT sensitive thermal flux sensors, and if we did it would not be a $10 experiment, but even a first year student can appreciate the first law of energy, and by simple application of logic that the Kelvin statement can be restated as "you cannot produce power from a device wholly immersed in a single thermal reservoir". So if you know the DUT is immersed in a single reservoir, which is in practical terms known to be so by making sure it has as close as possible to zero temp gradient (I can guarantee less than 1mK), then if it outputs electrical energy it must follow that it does so with 100% efficiency.


To prove satisfactory compliance to the issue of my oven being an isothermal reservoir I deliberately applied thermal gradients of 1deg K to the DUT, the output variance was less than 0.01%. I was therefore 100% sure that my results were valid and not simply a case of me heating a DUT with a temperature gradient.


Taking the view that I can reasonably estimate both the thermal flux through vacuum and the temperature depression of the DUT active elements (cathode and anode) against the electrical output, you get a value of less than 0.00001 Kelvin cathode to anode (do the calcs yourself), but lets increase it to 0.001K, then apply that to the carnot equation and you simply cannot get 4uW output, in fact if you care to do some calculations based on a DT of 1mK the result would not allow a Carnot limit output of any more than 4 pW, so I am very sure of what I say.


predicted carnot limit = 4 uW thermal input flux x .001/1000 = 4pW (this is 1,000,000 times smaller than the measure output)


So something exceeding the Carnot efficiency limit by such a massive factor should ring some positive bells in your head.


MarkE, let's agree that no matter what I say you will not agree, and you will not pursue it by doing your own experiment, and that accordingly this conversation has no point.


FWIIW I in fact posted the diagram that started this conversation for the benefit of profitis, I really did not want a debate with entrenched skeptics.


Have a nice day.


Bye all.


Phil

MarkE

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Re: quentron.com
« Reply #1990 on: February 03, 2014, 09:30:16 AM »

I am at a loss as to how you think, but anyway.


Of course it is going to be very difficult to show heat taken in by the DUT to electrical output unless we build around the DUT sensitive thermal flux sensors, and if we did it would not be a $10 experiment, but even a first year student can appreciate the first law of energy, and by simple application of logic that the Kelvin statement can be restated as "you cannot produce power from a device wholly immersed in a single thermal reservoir".

Unfortunately, you do not have a device immersed in a single thermal reservoir.  If you did, then in order to violate the Kelvin statement you would have to show that device dividing that single reservoir into hotter and colder sections without the benefit of outside energy.  On top of that you have a source of outside energy. 

We could take your experiment and instead of connecting the lead wires to a meter, enclose them in a cylinder fitted with a piston.  Heat conducted by the wires would heat and expand gas in the cylinder and push on the piston.  No thermodynamic laws would be violated. 
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 So if you know the DUT is immersed in a single reservoir, which is in practical terms known to be so by making sure it has as close as possible to zero temp gradient (I can guarantee less than 1mK), then if it outputs electrical energy it must follow that it does so with 100% efficiency.

That is not what you have.  See above.  If you have instrumentation capable of measuring to 0.001K accuracy at ~800K that works for less than $10. you can make lots of money in industrial temperature measurement.
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To prove satisfactory compliance to the issue of my oven being an isothermal reservoir I deliberately applied thermal gradients of 1deg K to the DUT, the output variance was less than 0.01%. I was therefore 100% sure that my results were valid and not simply a case of me heating a DUT with a temperature gradient.

Again, even if you could keep the temperature at various points in your fixture on an extreme knife's edge, the thermal leaks of the oven remain, including through your lead wires.  Steady temperature no more means no heat flow, than a constant voltage between two nodes of a circuit means no current flow through one or the other nodes.
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Taking the view that I can reasonably estimate both the thermal flux through vacuum and the temperature depression of the DUT active elements (cathode and anode) against the electrical output, you get a value of less than 0.00001 Kelvin cathode to anode (do the calcs yourself), but lets increase it to 0.001K, then apply that to the carnot equation and you simply cannot get 4uW output, in fact if you care to do some calculations based on a DT of 1mK the result would not allow a Carnot limit output of any more than 4 pW, so I am very sure of what I say.

This is called bootstrapping.  It is complete folly.  10E-6 K control out of 500K rise is 20 parts per billion net input / output power stability.  If you think such extraordinary power stability is available to a mains operated oven heater, then you are at serious odds with the state of the art.
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predicted carnot limit = 4 uW thermal input flux x .001/1000 = 4pW (this is 1,000,000 times smaller than the measure output)

In order for that 4uW to represent something that could challenge the Kelvin statement, the total rate of heat removal from the oven would also have to be 4uW.  If you are powering your oven with more than 4uW then the 4uW measurement does not help you challenge the Kelvin statement of the Second Law of Energy.
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So something exceeding the Carnot efficiency limit by such a massive factor should ring some positive bells in your head.

You can't show such a thing in the presence of thermal leaks all around you that are orders and orders of magnitude greater than the power you measure through your meter.
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MarkE, let's agree that no matter what I say you will not agree, and you will not pursue it by doing your own experiment, and that accordingly this conversation has no point.


FWIIW I in fact posted the diagram that started this conversation for the benefit of profitis, I really did not want a debate with entrenched skeptics.


Have a nice day.


Bye all.


Phil
If this is your experiment, it is no wonder that you cannot find traction with academic professionals.  At the low power levels you are talking about discerning power moved by what you think is a Second Law violating mechanism from power input by the oven is going to be quite tricky.  What you would be looking for is a way to make one part of the reservoir hotter, and one part colder.

wings

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Re: quentron.com
« Reply #1991 on: February 03, 2014, 09:37:28 AM »
"I have approximate answers and possible beliefs and different degrees of certainty about different things, but I'm not absolutely sure about anything."
~ Richard Feynman, Nobel Price in physics, 1965

MarkE

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Re: quentron.com
« Reply #1992 on: February 03, 2014, 09:47:47 AM »
Those are wise words from an extremely intelligent physicist.  Carefully designed experiments that can actually tell the difference between the null and the actual hypothesis are valuable tools that help us find our way through the darkness.  Experiments that can't discern the null from the actual hypothesis can't reveal anything new.  They can unfortunately mislead.  That seems to be the case with Mr. Hardcastle's experiment.  Whether his idea that the Second Law can be broken or not cannot be distinguished by his experiment.  Some other experiment is needed.

wings

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Re: quentron.com
« Reply #1993 on: February 03, 2014, 10:57:29 AM »
Second Law can be broken!

Another Way to Realize Maxwell’ s Demon: Xinyong Fu, Zitao Fu, Shanghai Jiao Tong University, September 28, 2005

The device can provide continuously a small but macroscopic
power to an external load, violating Kelvin’s statement of the second law.
http://arxiv.org/ftp/physics/papers/0509/0509111.pdf



http://arxiv.org/pdf/physics/0311104v3.pdf




http://physicsworld.com/cws/article/news/2012/mar/08/graphene-in-new-battery-breakthrough


MarkE

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Re: quentron.com
« Reply #1994 on: February 03, 2014, 12:06:08 PM »
Second Law can be broken!

Another Way to Realize Maxwell’ s Demon: Xinyong Fu, Zitao Fu, Shanghai Jiao Tong University, September 28, 2005

The device can provide continuously a small but macroscopic
power to an external load, violating Kelvin’s statement of the second law.
http://arxiv.org/ftp/physics/papers/0509/0509111.pdf



http://arxiv.org/pdf/physics/0311104v3.pdf




http://physicsworld.com/cws/article/news/2012/mar/08/graphene-in-new-battery-breakthrough
Maybe.  That is certainly what the claimants say.  In the first case, the video is instructional as to the types of experiments that they ran.  In each case a very powerful magnet was moved into place and/or rotated which induced an image current in the copper box.  The energy from that current can end up charging stray capacitance in the system.  The electrometer measures extremely small currents, so it could take some time to discharge.  The tests would be more convincing if they were run with the magnet in a fixed position relative to the copper box for longer periods of time, such as 24 hours.

The graphene "heat" battery uses two dissimilar metals in an electrolytic solution.  The researchers need to rule out that they have simply built an electrochemical battery.

If Mr. Hardcastle is to prove his claims, he needs to set-up a test that actually goes after the Kelvin statement of the Second Law of Energy.  The experiment that he has presented is not up to the task.