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Mechanical free energy devices => mechanic => Topic started by: allcanadian on January 25, 2015, 03:23:46 PM
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This is from a previous debate on another thread
Now your thinking, the DWFTTW process appeared to the weak minded as a violation of physics because you cannot get more than what is already there. However their error was in the Energy accounting of an open system and a failure to understand what was there and available to use. You are correct and the rules which apply to closed systems do not always apply to open systems because obviously they are not the same thing.
If I put 1w of electric energy into a resistance heater I will always get 1w of heat out but if I put 1w of electrical energy into a heat pump I may get 5w of heat out. A resistance heater is a closed system and a heat pump is an open system...it's that simple. The critics are simply arguing that all systems must remain closed, they are arguing that the simple resistance heater is the only way of doing things which as we know is pure delusion. Why if they had their way we would have to install our wind turbines, heat pumps and solar panels inside dark closed boxes just to satisfy their twisted notion of reality.
AC
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Your post is littered with false assertions answered in the other thread.
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Very true AC.
When a combination of a closed system and an open system are combined,we have a system that can deliver a net gain in energy output over that of which the closed system was supplied.The open system is ofcourse nature itself in the above case that AC mentioned,and the system i am about to present here over the next week or two. The time factor is due to the fact that i work long hours,and dont have a computor avaliable during working hours-the joys of being on the road for a living. The aditive to this time is that we will also !no doubt! have to deal with those in the know in regards to the concervation of energy ;)
Quote: In physics, the law of conservation of energy states that the total energy of an isolated system remains constant—it is said to be conserved over time. Energy can be neither created nor be destroyed, but it can change form, for instance chemical energy can be converted to kinetic energy in the explosion of a stick of dynamite.
A consequence of the law of conservation of energy is that a perpetual motion machine of the first kind cannot exist. That is to say, no system without an external energy supply can deliver an unlimited amount of energy to its surroundings.
All that is needed is an open system to supply the extra energy ontop of that which was supplied and accounted for in the closed system.
Our open system will consist of the gravitational force,and the buoyant force.
Gravitational force
Quote: gravitational force - (physics) the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface; "the more remote the body the less the gravity"; "the gravitation between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them";
The gravitational force will be supplying our system with a downward force,and that force is the mass of our system.
The buoyant force.
Quote: When an object is placed in a fluid, the fluid exerts an upward force we call the buoyant force. The buoyant force comes from the pressure exerted on the object by the fluid. Because the pressure increases as the depth increases, the pressure on the bottom of an object is always larger than the force on the top - hence the net upward force.
The buoyant force is what will supply the upward force in our system,and that force is dependant on the buoyant surface area,and the density of the mass within that buoyant surface area.
Below is a diagram of our closed system that conforms to the law of the conservation of energy-as stated in the diagram.
Our confirmation has been agreed upon-Quote: If the energy accounting balances, which everytime it has ever been done correctly it always has, then there is neither surplus nor deficit.
Quote:Energy is conserved in your electrolysis example.
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Both quoted statements are correct. Now feel free to try and generate a surplus or a deficit.
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"Energy can not be created or destroyed."
Do we always know where the energy goes in all possible cases? What I mean is...I believe that the above law is still intact. My question is, can we always know where energy is dissipated (as heat) or converted in any device? Is there any "gray" area here?
I have no examples at this time, but if I think about it, I may be able to come up with some possible scenarios.
I am not taking sides here, just wanting to learn.
Bill
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"Energy can not be created or destroyed."
Do we always know where the energy goes in all possible cases? What I mean is...I believe that the above law is still intact. My question is, can we always know where energy is dissipated (as heat) or converted in any device? Is there any "gray" area here?
I have no examples at this time, but if I think about it, I may be able to come up with some possible scenarios.
I am not taking sides here, just wanting to learn.
Bill
I am not aware of any such situations either.
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Both quoted statements are correct. Now feel free to try and generate a surplus or a deficit.
Are you saying that this system(the electrolisis unit) cannot or dose not have the capability to give rise to another form of energy output now that all energy within the system has been accounted for?
What if we were to change the enviroment in which this system opperates?
If we had a heat sink of sorts on the electrolisis unit to disipate the heat,would that change the outcome-->silly question i know,but needs to be asked none the less.
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Are you saying that this system(the electrolisis unit) cannot or dose not have the capability to give rise to another form of energy output now that all energy within the system has been accounted for?
I am saying that if one accounts for all of the energy/matter the total amount at the beginning and end of the process will be found to be the same.
What if we were to change the enviroment in which this system opperates?
What about it? Either you do all the accounting or you don't. If you do all the accounting correctly, then the energy/matter will balance.If we had a heat sink of sorts on the electrolisis unit to disipate the heat,would that change the outcome-->silly question i know,but needs to be asked none the less.
Add a heatsink, add a heater. It doesn't matter unless you fail to do the accounting correctly.
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I am saying that if one accounts for all of the energy/matter the total amount at the beginning and end of the process will be found to be the same.What about it? Either you do all the accounting or you don't. If you do all the accounting correctly, then the energy/matter will balance.Add a heatsink, add a heater. It doesn't matter unless you fail to do the accounting correctly.
OK-so we agree that we have accounted for all the energies within the electrolisis unit-correct?
You may find this a little !drag! MarkE,but piece by piece you are going to be the one that confirms this extra energy.
So in saying that,here is my next question.
Q-Dose it require energy to move a mass through a body of water?
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OK-so we agree that we have accounted for all the energies within the electrolisis unit-correct?
You may find this a little !drag! MarkE,but piece by piece you are going to be the one that confirms this extra energy.
So in saying that,here is my next question.
Q-Dose it require energy to move a mass through a body of water?
Not enough information. Move a mass through water in what direction? Up? Down (gravity) Left, right? How fast? How far? (time) I think I understand what you might be getting at here but, you probably need to re-phrase your question a little.
(Please notice I did not ask what the temperature was outside, or how large the body of water was, or, if it was salt water, or day, or night. So, I am not trying to be a smart ass.)
Bill
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Not enough information. Move a mass through water in what direction? Up? Down (gravity) Left, right? How fast? How far? (time) I think I understand what you might be getting at here but, you probably need to re-phrase your question a little.
(Please notice I did not ask what the temperature was outside, or how large the body of water was, or, if it was salt water, or day, or night. So, I am not trying to be a smart ass.)
Bill
Lol-your starting to sound like MH and MarkE-->or is that the intention.
Regardless-it dosnt matter/up down/left ./right-->it's moving a mass against an enviroment that has friction. When you move a mass through water,you displace the same volume of water as the volume of the mass. If a mass is sinking,then the force being applied to that mass is gravity-gravity is doing the work here. If the mass is rising,then the buoyant force is doing the work.
If the mass rises to the surface,it would have taken energy to get it to the bottom of the body of water. If the mass sink's,then it will take energy to get it to the surface.
But to top it off,we can move in any direction you like,it really dosnt matter ;). We can go left/right/up/down-->loop de loops and all :D
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Brad:
Thank you for clarifying that.
I have no answer so I will sit back and await replies from those more knowledgeable than myself.
Bill
PS Thank you for taking my question in the spirit in which it was intended.
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Brad:
Thank you for clarifying that.
I have no answer so I will sit back and await replies from those more knowledgeable than myself.
Bill
PS Thank you for taking my question in the spirit in which it was intended.
No prob
The answer is surprisingly slow comming.
Maybe careful thought is being given to this,as they know what's comming?.
Could the truck driver actually be able to show an extra energy output without it effecting the already accounted for energy input ???
Whats this,energy from gravity and buoyancy without any apparent energy input:D
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OK-so we agree that we have accounted for all the energies within the electrolisis unit-correct?
You may find this a little !drag! MarkE,but piece by piece you are going to be the one that confirms this extra energy.
So in saying that,here is my next question.
Q-Dose it require energy to move a mass through a body of water?
You may think so. You can play twenty questions all you want. When it comes down to performing an energy balance you will need to articulate a reasonably complete description of your set-up, measurement methods, and measurement results.
Q-Dose it require energy to move a mass through a body of water?
It takes energy to move a volume of anything, including water through a body of water just due to rheology.
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Lol-your starting to sound like MH and MarkE-->or is that the intention.
Regardless-it dosnt matter/up down/left ./right-->it's moving a mass against an enviroment that has friction. When you move a mass through water,you displace the same volume of water as the volume of the mass. If a mass is sinking,then the force being applied to that mass is gravity-gravity is doing the work here. If the mass is rising,then the buoyant force is doing the work.
The buoyant force is the force of gravity on the displaced volume of water. The water if it can will sink into the volume occupied by the buoyant float.If the mass rises to the surface,it would have taken energy to get it to the bottom of the body of water. If the mass sink's,then it will take energy to get it to the surface.
It takes work to increase the GPE of the system. When an object is sunk an equivalent volume of water must be lifted out of the space that the object ends up occupying. If the SG of the sunk object is less than that of the water it displaces, more mass moves up than moves down, requiring work. If the SG is greater than the surrounding water, then the object sinks passively, releasing work. Conversely, when a sunk object is lifted, a like volume of water falls into the space that the sunk object occupied. If the SG is less than the surrounding water then the water sinks passively, releasing work as the object rises. If the SG is greater than the surrounding water, then more mass moves up than down, and work must be done to lift the sunken object.
But to top it off,we can move in any direction you like,it really dosnt matter ;). We can go left/right/up/down-->loop de loops and all :D
If the SG is the same as the surrounding water then no GPE changes with any move. But rheology will still take its toll.
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No prob
The answer is surprisingly slow comming.
Maybe careful thought is being given to this,as they know what's comming?.
Could the truck driver actually be able to show an extra energy output without it effecting the already accounted for energy input ???
Whats this,energy from gravity and buoyancy without any apparent energy input:D
It's the same accounting problem as before.
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The buoyant force is the force of gravity on the displaced volume of water. The water if it can will sink into the volume occupied by the buoyant float.It takes work to increase the GPE of the system. When an object is sunk an equivalent volume of water must be lifted out of the space that the object ends up occupying. If the SG of the sunk object is less than that of the water it displaces, more mass moves up than moves down, requiring work. If the SG is greater than the surrounding water, then the object sinks passively, releasing work. Conversely, when a sunk object is lifted, a like volume of water falls into the space that the sunk object occupied. If the SG is less than the surrounding water then the water sinks passively, releasing work as the object rises. If the SG is greater than the surrounding water, then more mass moves up than down, and work must be done to lift the sunken object.If the SG is the same as the surrounding water then no GPE changes with any move. But rheology will still take its toll.
Truly excellent ;)
Piece by piece were getting it done Mark :)
Now one thing we forgot here to look at.
Is all the stored energy in our hydrogen and oxygen gases converted into heat energy if we ignite the mix of the two gases-->so we have mixed our two gases together,and we send that mix through say a hydroxy torch. We have ofcourse ignited the gas mix coming out of the torch.
We want to account for all the stored energy within the two gases now.
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Truly excellent ;)
Piece by piece were getting it done Mark :)
Now one thing we forgot here to look at.
Is all the stored energy in our hydrogen and oxygen gases converted into heat energy if we ignite the mix of the two gases-->so we have mixed our two gases together,and we send that mix through say a hydroxy torch. We have ofcourse ignited the gas mix coming out of the torch.
All the chemical bond energy will convert to heat in the reaction. When the resulting H2O falls back to the liquid state the phase change energy will also be released as heat. If you use a torch to ignite the mixture, then you could have a real measurement resolution issue to deal with, as you will need to know precisely how much heat the torch adds over the course of the event period.
We want to account for all the stored energy within the two gases now.
Yes we do.
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All the chemical bond energy will convert to heat in the reaction. When the resulting H2O falls back to the liquid state the phase change energy will also be released as heat.
Fabulous news Mark
If you use a torch to ignite the mixture, then you could have a real measurement resolution issue to deal with, as you will need to know precisely how much heat the torch adds over the course of the event period.
No No-the torch is what has the mixture of gas going through it-were not useing a torch to ignite the gas-we can use a small spark for that. we are running the HHO mix through the HHO torch as a way to mix and burn the gas, We could use any sort of gas heater designed to burn HHO to do the heating. As long as we have accounted for all the stored energy within the HHO mix being converted into heat during the phase change.
I would like you to look over the whole process now,and see that we have accounted for all the energy supplied by the battery as we feel we have. This energy supplied by the battery should now all have been converted into heat once the HHO has been ignited,and made that phase change back into water.
I am just about to hit the hay,and up early to start another weeks work on the road. I am going to give others also the chance to take a look over all the post here,and see if they think there is something we have missed.
We should now have accounted for all the energy that was supplied to the electrolisis system,and that supplied energy should now have been converted into heat.
Come friday night,i will put forth the open side of the design-->this is where the extra energy will come into play,but this is only after all here are happy with the summation so far.
P.S
Thanks for all your help on this Mark ;) I know some times we may not see eye to eye on what we believe to be true,but your help is much appreciated.
TinMan
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Fabulous news Mark
No No-the torch is what has the mixture of gas going through it-were not useing a torch to ignite the gas-we can use a small spark for that. we are running the HHO mix through the HHO torch as a way to mix and burn the gas, We could use any sort of gas heater designed to burn HHO to do the heating. As long as we have accounted for all the stored energy within the HHO mix being converted into heat during the phase change.
I would like you to look over the whole process now,and see that we have accounted for all the energy supplied by the battery as we feel we have. This energy supplied by the battery should now all have been converted into heat once the HHO has been ignited,and made that phase change back into water.
I am just about to hit the hay,and up early to start another weeks work on the road. I am going to give others also the chance to take a look over all the post here,and see if they think there is something we have missed.
We should now have accounted for all the energy that was supplied to the electrolisis system,and that supplied energy should now have been converted into heat.
Come friday night,i will put forth the open side of the design-->this is where the extra energy will come into play,but this is only after all here are happy with the summation so far.
P.S
Thanks for all your help on this Mark ;) I know some times we may not see eye to eye on what we believe to be true,but your help is much appreciated.
TinMan
We have talked about the different things going on. We have yet to perform any kind of numerical accounting. An accounting means that we will actual construct a model even if it is just a simple paper form or better a spreadsheet. Then there is the matter of obtaining data to assign each item.
Is it your belief that you are going to get a surplus of energy on paper without performing measurements?
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We have talked about the different things going on. We have yet to perform any kind of numerical accounting. An accounting means that we will actual construct a model even if it is just a simple paper form or better a spreadsheet. Then there is the matter of obtaining data to assign each item.
Is it your belief that you are going to get a surplus of energy on paper without performing measurements?
This is why you are doing all the accounting for the energy conversions Mark-you are putting it on paper. We are useing your beloved laws of physics to account for all energy in,and all energy out/disipated. This is why i am asking you(and others) to make sure we have converted and accounted for all the energy the battery delivered.
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This is why you are doing all the accounting for the energy conversions Mark-you are putting it on paper. We are useing your beloved laws of physics to account for all energy in,and all energy out/disipated. This is why i am asking you(and others) to make sure we have converted and accounted for all the energy the battery delivered.
In order to do so we have to agree on all the inputs and outputs. When you are ready to describe the entire test set-up, we can do that.
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"HHO" implies monoatomic gases. These gases (hydrogen and oxygen) do not exist in unbound monoatomic forms at normal temperatures and pressures! They exist as diatomic molecules, H2 and O2. Careful measurements of volumes, and considering the combined Boyle-Charles-GayLussac law, will show the true nature of the evolved gases.
From one mole of water, 18 grams of H2O, "HHO" implies 3 moles (~22.4 liters x 3) of evolved dry gases at STP. But if the gases are actually the diatomic species H2 and O2 we expect one mole of H2 and one-half mole of O2 from one mole of H2O, that is (~22.4 liters x 1.5) of evolved dry gasses at STP.
Has anyone who claims "HHO" monoatomic gases actually done this molar volume measurement properly (that is, drying and cooling the evolved gases to make dry gas at STP for the volume measurement)?
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This gas buoancy device will work.h2 gas is swallowed at cathode and spits out of anode.only 0.1v-0.2v power source is needed and height of column can be made arbritarily large.this will work because entropy actually increases at base and decreases at top making some margin for gain
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In order to do so we have to agree on all the inputs and outputs. When you are ready to describe the entire test set-up, we can do that.
There will be no more input to the electrolisis unit other than that of what the batteries supply.There will however be an additional output from an open system that is a result of the actions of the closed system-but dose not impinge on that closed system-the electrolisis system.
If you are happy that we have accounted for all the energy the battery supplied to the electrolisis unit in the end form of heat, then we will move on. If you are not satisfied that we have converted all the supplied energy into heat, then this must be resolved before we go any further. The reason being that it must be a proof of concept on paper, as it would be very expensive to build and test-but with the monies avaliable, it can be done quite easily
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There will be no more input to the electrolisis unit other than that of what the batteries supply.There will however be an additional output from an open system that is a result of the actions of the closed system-but dose not impinge on that closed system-the electrolisis system.
If you are happy that we have accounted for all the energy the battery supplied to the electrolisis unit in the end form of heat, then we will move on. If you are not satisfied that we have converted all the supplied energy into heat, then this must be resolved before we go any further. The reason being that it must be a proof of concept on paper, as it would be very expensive to build and test-but with the monies avaliable, it can be done quite easily
Please draw a simple sketch of where you want to define the boundaries of your inputs and outputs. We have the battery that will apply some constant amount of power for a fixed amount of time? We have the water volume at an initial temperature and volume? We have an internal gas volume at an intial temperture and volume? We are then going to run the unit and do what with the gas as it emanates? Do you intend to let it escape and then ignite it? Or do you intend to collect it at some pressure and measure the volume under an assumption of an HHV value? What do you plan to do with the water that results? Do you want to return it to the vessel? Finally, whatever this open system you wish to attach to, you need to define what will be exchanged with it, so that we can measure it, whether we do that hypothetically on paper or in an actual experiment.
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Please draw a simple sketch of where you want to define the boundaries of your inputs and outputs. We have the battery that will apply some constant amount of power for a fixed amount of time? We have the water volume at an initial temperature and volume? We have an internal gas volume at an intial temperture and volume? We are then going to run the unit and do what with the gas as it emanates? Do you intend to let it escape and then ignite it? Or do you intend to collect it at some pressure and measure the volume under an assumption of an HHV value? What do you plan to do with the water that results? Do you want to return it to the vessel? Finally, whatever this open system you wish to attach to, you need to define what will be exchanged with it, so that we can measure it, whether we do that hypothetically on paper or in an actual experiment.
The gas will be stored at pressure, then once we are ready, we will ignite the gas mix, and theresulting water will be returned back to the electrolisis cell for use on the second cycle.
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Is the intent to keep the cell highly insulated or exchange the heat evolved to an external sink?
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Is the intent to keep the cell highly insulated or exchange the heat evolved to an external sink?
We want to heat a sealed vessel that the complete electrolisis unit is housed in. I needed a way to heat a sealed area (room so to speak), but also a force to opperate the workings of the secondary open system. That force is the pressure in the storage tanks for the H and O. This pressure would be there regardless if the secondary system is there or not. This is why the secondary system is an open system, as it dose not impinge on the primary system in any way.
The primary system effects the secondary open system, but the secondary system dose not effect the primary system.
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We want to heat a sealed vessel that the complete electrolisis unit is housed in. I needed a way to heat a sealed area (room so to speak), but also a force to opperate the workings of the secondary open system. That force is the pressure in the storage tanks for the H and O. This pressure would be there regardless if the secondary system is there or not. This is why the secondary system is an open system, as it dose not impinge on the primary system in any way.
The primary system effects the secondary open system, but the secondary system dose not effect the primary system.
When you've settled in on what you want please draw a sketch.
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When you've settled in on what you want please draw a sketch.
A sketch will isnt possable while im on the road.
All you have to picture is the complete system (as per diagram posted on first page of this thread) in an insulated room.
We are just confirming that all enervy has been accounted for in the form of heat. Like I said, the secondary system dose not effect the primary system.
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@MarkE
At this point in time-now we have agreed that energy is conserved in the electrolisis system, can you see any way of returning energy back to the battery with out it effecting the output ofthe system at all?.
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This is from a previous debate on another thread
Now your thinking, the DWFTTW process appeared to the weak minded as a violation of physics because you cannot get more than what is already there. However their error was in the Energy accounting of an open system and a failure to understand what was there and available to use. You are correct and the rules which apply to closed systems do not always apply to open systems because obviously they are not the same thing.
If I put 1w of electric energy into a resistance heater I will always get 1w of heat out but if I put 1w of electrical energy into a heat pump I may get 5w of heat out. A resistance heater is a closed system and a heat pump is an open system...it's that simple. The critics are simply arguing that all systems must remain closed, they are arguing that the simple resistance heater is the only way of doing things which as we know is pure delusion. Why if they had their way we would have to install our wind turbines, heat pumps and solar panels inside dark closed boxes just to satisfy their twisted notion of reality.
AC
A resistive heater is not a closed system. That is obvious because energy leaves the system, if it was closed no energy could enter or leave the system. Very simple.
In reality there is only one closed system which is the entire Universe (meaning everything that is) within the Universe there is no "real" closed systems, however we can consider an almost closed system as a closed system based on input alone and disregard the energy leaving open the system if we want, but what does that achieve.
Any system that allows energy to enter or Leave the system is open by definition.
Anything with an output or an input is an open system. Generally we put energy into a system to get energy out, regardless of the ratios they are all open systems. A closed system is useless because there can be no output by definition, unless your are contained within it that is, then it can be useful to you.
Cheers.
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Wether a system is open or closed has nothing to do with the propensity toward thermodynamics laws violations.spontaneous concentration of heat is itself banned regardless
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Mmm
Mark has seemed to have disappeared.
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@Farmhand
A resistive heater is not a closed system. That is obvious because energy leaves the system, if it was closed no energy could enter or leave the system. Very simple.
I think you know exactly what I meant and squabbling over semantics is a losing proposition.
@profitis
Thermodynamics is just another way of saying I do not understand modern physics in my opinion. Fundamentally there is no such thing as "heat" as most people understand it. Heat is a measure of the kinetic energy of particles and it also relates to a narrow band of wavelengths within the EM spectrum. Thus Heat is fundamentally an electrodynamic phenomena as most all phenomena are. In my opinion when people quote thermodynamics they are basically saying--- hey look I'm still stuck in the 1900's way of thinking.
In the modern world scientists are converting this kinetic and EM energy directly into electrical energy with nano-conductors and other engineered materials. In fact many people here might accuse these scientists of making a magical perpetual motion machine as it has no moving parts and could be stuck inside a black box...imagine that.
In fact we may even go so far as to say these new engineered materials are equivalent to Maxwell's Demon as imagined by him a very long time ago.
AC
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Moving forward... I understand almost everyone is completely preoccupied by entropy ie thermodynamics and in a sense we could call this energy source technology. That is we transform energy in a material into heat energy and in many cases apply this heat energy to an archaic and outdated system to produce electrical energy. We basically spew most of the energy into the environment never to return and we call this modern.
However whenever I mention an energy sink technology nobody has a clue what I'm talking about and give me this puzzled look as if they have just seen a unicorn. Energy sink technology?, What do you mean... Energy sink technology. In fact the thought had never even crossed there mind that all that energy radiating away from their source might actually go somewhere.
My first exposure to this concept was when I read Nikola Tesla's lecture "The problem of increasing human energy". http://www.tfcbooks.com/tesla/1900-06-00.htm (http://www.tfcbooks.com/tesla/1900-06-00.htm)
DIAGRAM b. OBTAINING ENERGY FROM THE AMBIENT MEDIUM
But was it not possible to realize a similar condition without necessarily going to a height? Conceive, for the sake of illustration, [a cylindrical] enclosure T, as illustrated in diagram b, such that energy could not be transferred across it except through a channel or path O, and that, by some means or other, in this enclosure a medium were maintained which would have little energy, and that on the outer side of the same there would be the ordinary ambient medium with much energy. Under these assumptions the energy would flow through the path O, as indicated by the arrow, and might then be converted on its passage into some other form of energy. The question was, Could such a condition be attained? Could we produce artificially such a "sink" for the energy of the ambient medium to flow in? Suppose that an extremely low temperature could be maintained by some process in a given space; the surrounding medium would then be compelled to give off heat, which could be converted into mechanical or other form of energy, and utilized. By realizing such a plan, we should be enabled to get at any point of the globe a continuous supply of energy, day and night. More than this, reasoning in the abstract, it would seem possible to cause a quick circulation of the medium, and thus draw the energy at a very rapid rate.
As is often the case Tesla was light years ahead of his time and you guys can read whatever you like into this lecture. In essence this is what everyone is looking for... An Energy Sink.
AC
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Haha, I'm not arguing, there is nothing to argue a resistive heater is not a closed system. True or False ?
You over complicate things to try to look clever. When a lizard basks in the sun it is an energy sink. Clearly Free energy for the
lizard. Solar panel is similar.
Thermo electric modules ect. can be energy sinks. You must be talking to very unimaginative people.
EDIT: OK I see Tesla's Quote. disregard the obsolete parts of the following paragraph. My bad.
Care to explain your revolutionary energy sink idea, I know you will never show anything so how about an explanation to explain
the working concept ? Do you mean a "sink" like a heat sink or a "sink" like water flowing down the drainage portion of a kitchen
sink, but an energy flow down the sink rather than the water analogy ?
I think a modern heat pump achieves that end, it sinks energy from the surrounding medium.
I'm not trying to be problematic, I'm always interested in interesting idea's.
There is nothing to win or lose. Only trying to save some young fella from going to his teacher and claiming a resistive heater is a
closed system.
Show me a truly closed system on a bench, one where no energy can enter or leave the system. Take a picture of it.
..
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Sure sure @allcanadian.there's just one problem: kelvin statement and entropy laws are still written in todays textbooks in the same old wording as they were 100yrs ago so I stay with such outdated conceptual visualization for the sake of the kids here who have been brainwashed by it since birth.few people are going to understand phonon and photon resonant up-conversion
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yes Mark E does seem to have taken a break.....
hope all is well ?
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Haha, I'm not arguing, there is nothing to argue a resistive heater is not a closed system. True or False ?
You over complicate things to try to look clever. When a lizard basks in the sun it is an energy sink. Clearly Free energy for the
lizard. Solar panel is similar.
Thermo electric modules ect. can be energy sinks. You must be talking to very unimaginative people.
EDIT: OK I see Tesla's Quote. disregard the obsolete parts of the following paragraph. My bad.
Care to explain your revolutionary energy sink idea, I know you will never show anything so how about an explanation to explain
the working concept ? Do you mean a "sink" like a heat sink or a "sink" like water flowing down the drainage portion of a kitchen
sink, but an energy flow down the sink rather than the water analogy ?
I think a modern heat pump achieves that end, it sinks energy from the surrounding medium.
I'm not trying to be problematic, I'm always interested in interesting idea's.
There is nothing to win or lose. Only trying to save some young fella from going to his teacher and claiming a resistive heater is a
closed system.
Show me a truly closed system on a bench, one where no energy can enter or leave the system. Take a picture of it.
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A closed system is one where the output dose not reach the enviroment. As this systems sole purpose is to create heat, and the system is housed in an insulated vessel, then no heat reaches the enviroment -thus the system is closed off from the enviroment. An open system is one that can use enviromental resources when there there to use.
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HHMmmm
This is a test of the Emergency Mark E contact system...
I spoke with Mr.Potter this AM [GDS] I have a very important update.
Mark E you can get this update at Chetkremens@gmail.com
please be aware if you have been Moderated ....I had nothing at all to do with that.
Chester Don't play like that !
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@Farmhand
I think I jumped the gun in my responses and sometimes I can be overly critical trying to justify another perspective. My bad
In the fewest words possible, nature transforms energy into other forms which are not opposed by the original form entering the system. Nature absorbs oranges and spits out apples which is unlike anything we do. We simply pump the oranges from one place to another and we always lose.
Think of a tree, we plant a tiny seed weighing grams and years later it is a huge tree weighing tons and all the energy to do this was absorbed from the environment. What we do is burn the damn tree rather than trying to understand how it works and how to copy these natural processes which is called Biomimicry.
AC
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The energy-sink perspective is a nice one though as that is literally all it is,just a kink-flow of energy in the middle of the room.
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@profitis
The energy-sink perspective is a nice one though as that is literally all it is,just a kink-flow of energy in the middle of the room.
Maybe you could explain your theory to the best scientists in the best universities developing engineered materials like nano-conductors and multi-layered multi-spectrum arrays. You see the science journals and the investors clamoring to invest hundreds of millions of dollars in this technology seem to think you are completely wrong. The science journals are saying this is a leap in technology like nothing we have ever seen in the history of mankind.
Who to believe, the best and the brightest on the cutting edge of technology or some guy in a forum who seems to be stuck in the past?. I find it very odd that one person in one place can believe one thing and yet in another place another person is changing the world and receiving praise from their peers while investors throw money at them like it's going out of style. How one person can be successful beyond our wildest dreams while others sit in a forum telling everyone it cannot be done. The reality of it boggles the mind.
AC
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sewer forums can also be the launchpad to fame and riches mr canadian.what you give you can get.
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@profitis
I'm just saying what we should already know on some level. There are a great deal of very smart young people doing unbelievable things which we thought were impossible. That's the way life is and all the old people always say it cannot be done while the younger generation are always busy doing it. You know we have come a long way but some things never seem to change.
AC
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Too true mr canadian.I'm not young nor old but if I was living where you are I probly wouldve been rich ages ago.here in africa if you don't bribe some-one in the right way you get nowhere.wild west(south) out here
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A closed system is one where the output dose not reach the enviroment. As this systems sole purpose is to create heat, and the system is housed in an insulated vessel, then no heat reaches the enviroment -thus the system is closed off from the enviroment. An open system is one that can use enviromental resources when there there to use.
Really ? no heat or energy at all can escape ? And no energy can enter either. How do you measure anything in a truly closed system
unless your inside it.
That's your definition.
An open system is one that can use enviromental resources when there there to use
My definition of a closed system is one where no energy can enter or escape, the energy within the system remains the same. Hence no closed system is of use to anyone unless your in it.
All our man made systems are open to some degree.
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Well even though Mark seem's to have vacated the premises,we will press on.
This is the first system(a proof of concept) to show how the hydroxy gas can be made to do useful work without any energy input needed-->energy is conserved within the system.
In the diagram below,you can see that we are storing the gas produced by the electrolisis system in a storage tank. Once the pressure reaches say 100psi within that tank,we have transformed some of the energy delivered by the battery into heat,and the rest of the energy delivered by the battery is stored within the hydroxy gas-->energy has been concerved at this point. Once the tank reaches 100psi,valve A opens,and the piston in the pneumatic ram is forced to the opposite end of the ram.This is where we can gain useful work at NO cost to the system,as the energy is still stored within the gas. When the piston in the pneumatic ram reaches it's travel limit,valve A closes,and valve B open's. The spring within the ram pushes the piston back up,and the hydroxy gas within that pneumatic ram is sent to the hydroxy heater(a gas heater converted to run on hydroxy). The energy within the hydroxy gas is converted into heat energy,and once again,energy has been conserved. After one full cycle,the heat energy out is equal to the electrical energy supplied by the battery,and no energy was consumed to gain useful work from the pneumatic cylinder.
This is one (small scale) example of how hydroxy gas can be used to gain energy within a closed system-->the total energy out is greater than that of which was supplied to the system.
Hydroxy(HHO) gas is used as it stores energy within it that can be reclaimed after the gas under pressure has done useful work.
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Mmm
Mark has seemed to have disappeared.
I've been on the road for three days.
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Well even though Mark seem's to have vacated the premises,we will press on.
This is the first system(a proof of concept) to show how the hydroxy gas can be made to do useful work without any energy input needed-->energy is conserved within the system.
In the diagram below,you can see that we are storing the gas produced by the electrolisis system in a storage tank. Once the pressure reaches say 100psi within that tank,we have transformed some of the energy delivered by the battery into heat,and the rest of the energy delivered by the battery is stored within the hydroxy gas-->energy has been concerved at this point.
The fallacy is in assumptions about the distribution of energy from the battery. The amount of energy that goes into the phase change increases with the ambient pressure. less and less gas is produced by each J from the battery as the ambient pressure increases.Once the tank reaches 100psi,valve A opens,and the piston in the pneumatic ram is forced to the opposite end of the ram.This is where we can gain useful work at NO cost to the system,as the energy is still stored within the gas.
Nope. You just expended energy from the battery.[qote] When the piston in the pneumatic ram reaches it's travel limit,valve A closes,and valve B open's. The spring within the ram pushes the piston back up,and the hydroxy gas within that pneumatic ram is sent to the hydroxy heater(a gas heater converted to run on hydroxy). The energy within the hydroxy gas is converted into heat energy,and once again,energy has been conserved. After one full cycle,the heat energy out is equal to the electrical energy supplied by the battery,and no energy was consumed to gain useful work from the pneumatic cylinder.[/quote]Nope. As stated before the fallacy is the incorrect belief that the rate of gas production is independent of the surrounding pressure.
This is one (small scale) example of how hydroxy gas can be used to gain energy within a closed system-->the total energy out is greater than that of which was supplied to the system.
Hydroxy(HHO) gas is used as it stores energy within it that can be reclaimed after the gas under pressure has done useful work.
It is yet another example of how a false assumption leads to an erroneous conclusion.
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I Think tinman means something like this.extra work out is obtained via density changes in this case.column is arbritarily tall.you may have to include pulser along the circuit
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The fallacy is in assumptions about the distribution of energy from the battery. The amount of energy that goes into the phase change increases with the ambient pressure. less and less gas is produced by each J from the battery as the ambient pressure increases.
Once again Mark you are missinterpreting my post. I made no assumptions at all.
The higher the pressure,the more compressed the bubbles of gas are forming on each plate. Each plate will continue to produce gas until such time as the pressure reaches the point as to when the gas becomes liquid-and we are no where near that point.
So i leave you to think about these two posabilities,and ponder as too which would be true.
1-the output of gas production remains the same,as the compressed gas= ambiant pressure amount.
2-The lower gas production due to rising pressure=a higher heat production by the cell.
One or a combination of the two are true-->but energy IS conserved.
Nope. You just expended energy from the battery
Incorrect,as the energy has already been accounted for in the production of heat,and energy stored within the gas-->as you have accepted as being true already. Storeing the gas at pressure dose not change the ratio of energy used(input) to the energy converted into heat and stored in the hydroxy gas.
.Nope. As stated before the fallacy is the incorrect belief that the rate of gas production is independent of the surrounding pressure.It is yet another example of how a false assumption leads to an erroneous conclusion.
Please feel free to show us where energy is destroyed,and not conserved.
As i stated,producing and storing the gas at pressure dose NOT change the energy output of the system. If less gas is produced when under pressure,then more heat is produced as a result-->energy is conserved.
The pneumatic cylinder makes no impact on the energy stored in the gas,as the amount of gas remains the same regardless of wether the cylinder is there or not. The rest of the energy was converted into heat during the electrolisis process.
I am afraid you are wrong this time Mark-but please feel free to prove me incorrect.
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When the gas reaches a pressure of 120psi(there abouts,depending on tank size),and we are able to supply the pneumatic cylinder with a constant 100psi,it will give an equivalent energy output of 8728 joules of energy at no cost to the system.
As has been agreed upon,energy within the electrolisis system is conserved,and this conservation of energy dose not change due to either a pressure increase or decrease within the gas storage system. A higher pressure results in less gas production per joule of energy,but at the same time it results in a higher heat output from the electrolisis system-->energy must be conserved-->the laws of physics states this to be true.
It is also true that by increasing the storage tank volume and decreasing the pressure of gas within that storage tank,dose not decrease the energy stored within that gas-as long as the same amount of gas remains within that storage medium.
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Once again Mark you are missinterpreting my post. I made no assumptions at all.
The higher the pressure,the more compressed the bubbles of gas are forming on each plate. Each plate will continue to produce gas until such time as the pressure reaches the point as to when the gas becomes liquid-and we are no where near that point.
The erroneous assumption that you made and you continue to make is that the gas production rate in moles/Joule is constant independent of the external pressure. If you don't believe me then construct an experiment. So i leave you to think about these two posabilities,and ponder as too which would be true.
1-the output of gas production remains the same,as the compressed gas= ambiant pressure amount.
2-The lower gas production due to rising pressure=a higher heat production by the cell.
One or a combination of the two are true-->but energy IS conserved.
Incorrect,as the energy has already been accounted for in the production of heat,and energy stored within the gas-->as you have accepted as being true already. Storeing the gas at pressure dose not change the ratio of energy used(input) to the energy converted into heat and stored in the hydroxy gas.
You have led yourself down a garden path. The gas generation rate in molecules / Joule, IE moles / Joule depends is limited by the net available energy B E C A U S E as we have both agreed: ENERGY IS CONSERVED. You have agreed that a portion of the input energy goes into the phase change from liquid to gas. Now it is down to how much energy. The amount of energy required depends on the pressure and temperature. Increase the pressure as you have and you indentically increase the work required to evolve a given number of gas molecules. The energy stored in the compressed gas volume identically comes from the battery energy source, and idnetically takes away from the enrgy available electrolyze additional water molecules.
If your idea were correct, then we could destroy energy by pumping down your cylinder with a vacuum pump. Then we would have to add external work to move your piston while according to your idea evolving the same number of gas molecules per Joule of input battery energy as at 1 ATM or ~7 ATMs as you propose. Have you ever travelled to the mountains?
Did you ever notice that water boils faster the higher the altitude, IE the lower the ambient pressure? The same principles are at work here. You are "boiling" the water by electrolyzing it. At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure.
.Please feel free to show us where energy is destroyed,and not conserved.
You are unfortunately the one doing the funny accounting.As i stated,producing and storing the gas at pressure dose NOT change the energy output of the system. If less gas is produced when under pressure,then more heat is produced as a result-->energy is conserved.
Less gas is produced. An identically reduced amount of heat is evolved during the subsequent combustion. The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule. Again, if you do not believe me, do the math or do an experiment.The pneumatic cylinder makes no impact on the energy stored in the gas,as the amount of gas remains the same regardless of wether the cylinder is there or not. The rest of the energy was converted into heat during the electrolisis process.
This is again wrong. You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions. This is first year college chemistry.I am afraid you are wrong this time Mark-but please feel free to prove me incorrect.
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Quote frm markE:' Increase the pressure as you have and you indentically increase the work required to get a given amount of gas molecules'
Unquote
True.this is found by substituting pressure(p) for concentration (c) in the nernst equasion.HOWEVER,you forgot something very important: conversion of liquid to gas cools the liquid down and increases dramaticaly its entropy.this increase in entropy more (much more) compensates for the losses incurred on the nernst pressure joules.you gain,if you do it right
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Put simply: h2 + o = h2o + energy - gas to liquid phase change energy.. and h2o + energy = h2 + o + liquid to gas phase change energy.phase change energy are direct as usable heat to and from environment at no cost.E(gas electrode)= RT/nf ln P.increase P and you decrease T at the same time from phase change liquid to gas.
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At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure. You are unfortunately the one doing the funny accounting.Less gas is produced. An identically reduced amount of heat is evolved during the subsequent combustion. The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule. Again, if you do not believe me, do the math or do an experiment.This is again wrong. You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions. This is first year college chemistry.
The erroneous assumption that you made and you continue to make is that the gas production rate in moles/Joule is constant independent of the external pressure.
I made no such assumption-->once again this is yet another faulse claim you make. The loss in gas production per joule due to pressure increase is made up for in the way of heat output energy by the system.
You have agreed that a portion of the input energy goes into the phase change from liquid to gas. Now it is down to how much energy.
The energy that went into the phase change from liquid to gas will be the same as the energy returned when that phase change is reversed from gas to liquid-->unless you have found a way to destroy energy?.
The amount of energy required depends on the pressure and temperature. Increase the pressure as you have and you indentically increase the work required to evolve a given number of gas molecules.
That is correct,and that increase in work results in a higher heat output,and a lower number of gas molecules-->once again,energy is conserved.
Did you ever notice that water boils faster the higher the altitude, IE the lower the ambient pressure? The same principles are at work here. You are "boiling" the water by electrolyzing it.
Now i think you are taking the trip up the garden path. We are not boiling the water by electrolyzing it.. We are dissociating the water into H2 and O2. I have no idea as to how you think this is steam-->which is the result of boiling water.
The energy stored in the compressed gas volume identically comes from the battery energy source, and idnetically takes away from the enrgy available electrolyze additional water molecules.
That is right,the energy stored within the gas is the exact amount that was drawn from the battery to create it-->the rest of the energy supplied by the battery during this process is transformed into heat.. So please explain how reducing the pressure of that stored gas by enlarging the storage device changes any of the energy accounting after the gas has been produced.
If your idea were correct, then we could destroy energy by pumping down your cylinder with a vacuum pump.
Once again,another incorrect statement. To pump down the cylinder requires work to be done,and thus energy is required to do so. You are now transforming energy,not destrying it.
Then we would have to add external work to move your piston while according to your idea evolving the same number of gas molecules per Joule of input battery energy as at 1 ATM or ~7 ATMs as you propose.
And again-another faulse accusation. I never said the same number of gas molecules are produced per joule of input energy as the storage pressure go's up. I clearly stated that more energy from the battery would be transformed to heat,and less to gas production as the pressure builds within the storage medium-->energy is conserved.\
Have you ever travelled to the mountains?
We dont have mountains here-->are you not happy with YOUR garden path?
How you ever came about the idea that by increasing the size of the storage vessel ,and inturn decreasing the gas pressure uses or depletes the stored energy of the gas-- >is way beyond the garden path Mark.
Remove the pneumatic ram from the equasion,so as we just have our storage tank. We produce the gas until such time as that pressure in the tank reaches 120psi gage pressure. Energy has been conserved-->the amount of energy depleted from the battery is the exact amount of heat energy disipated by the circuit as a whole,and stored within the gas-->we have NOT destroyed any energy.
Now that all the energy has been accounted for,your saying that !some how! we are going to loose some of the stored energy by increasing the tank size,and decreasing the pressure of the stored gas. You MarkE will be the first to destroy energy.
You say that this is basic math,and yet you have screw'd it up-->is google not working right for you ATM?. :o
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At a lower ambient pressure it takes less work to do the "boiling" part that than it does at high ambient pressure. You are unfortunately the one doing the funny accounting.Less gas is produced. An identically reduced amount of heat is evolved during the subsequent combustion. The adiabatic heating increases which further decreases the gas generation rate in absolute molecules / Joule. Again, if you do not believe me, do the math or do an experiment.This is again wrong. You have drawn your boundaries incorrectly resulting in incorrect accounting and conclusions. This is first year college chemistry.
I made no such assumption-->once again this is yet another faulse claim you make. The loss in gas production per joule due to pressure increase is made up for in the way of heat output energy by the system.
The energy that went into the phase change from liquid to gas will be the same as the energy returned when that phase change is reversed from gas to liquid-->unless you have found a way to destroy energy?.
Of course not, but your invalid claims are tantamount to being able to both create and destroy energy: Simply change the ambient pressure. And that is of course absurd.
That is correct,and that increase in work results in a higher heat output,and a lower number of gas molecules-->once again,energy is conserved.
Energy is completely conserved all the way through. There is no surplus for free work from operating the piston, just as no deficit could be created by drawing a vacuum.
Now i think you are taking the trip up the garden path. We are not boiling the water by electrolyzing it.. We are dissociating the water into H2 and O2. I have no idea as to how you think this is steam-->which is the result of boiling water.
"Boiling" note the quotes refers to the liquid to gas phase change. That change requires less energy into a lower pressure atmosphere than into a higher pressure atmosphere. That difference leaves more or less energy to break chemical bonds electrolyzing the water. Higher pressure atmosphere => lower number of molecules of H2 and O2 evolved. If you don't believe me then build your apparatus and attempt to wow the world.That is right,the energy stored within the gas is the exact amount that was drawn from the battery to create it-->the rest of the energy supplied by the battery during this process is transformed into heat..
There you go with the screwy accounting again. The enrgy from the battery goes into: chemical bond splitting, and the phase change. That phase change results in stored energy in latent heat of vaporization and the pressure * volume product. When you remove some of that work from the system by way of the moving piston, it has left the building, Elvis is on stage no more and the amount of energy that you can recover from what remains is now diminished. The accounts remain balanced, and your scheme has failed to produce the excess energy you claim. So please explain how reducing the pressure of that stored gas by enlarging the storage device changes any of the energy accounting after the gas has been produced.
It is not ex facto. It is intrinsic to the phase change.Once again,another incorrect statement. To pump down the cylinder requires work to be done,and thus energy is required to do so. You are now transforming energy,not destrying it.
You are simply and utterly wrong. Take any volume of gas, and reduce the pressure, IE remove gas, and the remaining energy in the volume has been reduced: Ideal Gas Law, Charles Law, and Boyle's Law all apply.And again-another faulse accusation. I never said the same number of gas molecules are produced per joule of input energy as the storage pressure go's up. I clearly stated that more energy from the battery would be transformed to heat,and less to gas production as the pressure builds within the storage medium-->energy is conserved.\
Yes energy is conserved and no surplus becomes available because of the extra pressure and no extra can be extracted by operating the cylinder. Only by applying funny accounting do you challenge those facts.We dont have mountains here-->are you not happy with YOUR garden path?
I am not joining you on your stroll down yours.
How you ever came about the idea that by increasing the size of the storage vessel ,and inturn decreasing the gas pressure uses or depletes the stored energy of the gas-- >is way beyond the garden path Mark.
Actually do the math or do the experiment and maybe you will see past your false assumptions.Remove the pneumatic ram from the equasion,so as we just have our storage tank. We produce the gas until such time as that pressure in the tank reaches 120psi gage pressure. Energy has been conserved-->the amount of energy depleted from the battery is the exact amount of heat energy disipated by the circuit as a whole,and stored within the gas-->we have NOT destroyed any energy.
Now you're thinking. And for those exact reasons your proposed scheme fails.Now that all the energy has been accounted for,your saying that !some how! we are going to loose some of the stored energy by increasing the tank size,and decreasing the pressure of the stored gas. You MarkE will be the first to destroy energy.
Oops, now you've jumped right on your garden path again.
You say that this is basic math,and yet you have screw'd it up-->is google not working right for you ATM?. :o
Obviously your hand calculator is not working for you. Neither is basic logic.
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Mark:
Good to see that you are back and in good health.
Bill
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Mark:
Good to see that you are back and in good health.
Bill
Bill thanks. I was out of town. A friend of mine who was at the same event got food poisoning Tuesday. He seemed fine at his presentation on Wednesday.
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Of course not, but your invalid claims are tantamount to being able to both create and destroy energy: Simply change the ambient pressure. And that is of course absurd.
Lol,this is funny. Please show where the energy was created and destroyed.
Energy is completely conserved all the way through. There is no surplus for free work from operating the piston
Incorrect,as the energy has already been accounted for before the valve opens to opperate the piston in the ram.
That change requires less energy into a lower pressure atmosphere than into a higher pressure atmosphere.
Indeed when we are talking gas production. But in a higher pressure atmosphere where less gas is being produced per joule of energy,more heat is produced by the circuit as a whole,and once again,energy is conserved.
That difference leaves more or less energy to break chemical bonds electrolyzing the water. Higher pressure atmosphere => lower number of molecules of H2 and O2 evolved. If you don't believe me then build your apparatus and attempt to wow the world.
Unlike you infer,i have never disagreed with this. But when this is true,it is also true that the heat energy disipated by the circuit as a whole,is equal to the difference in energy supplied by the battery,and energy stored within the hydroxy gas.Once again,energy is conserved.
That phase change results in stored energy in latent heat of vaporization and the pressure * volume product.
No,a lot of the heat is a result or resistive losses,and chemical reactions within the battery itself.
These are things you should know mark :o I think maybe it is you who may need some bench time here to wow the world. Take an AC current and pass that through an Elite-no gas production,but heat is produced via the resistance within the Elite. To say,Quote: The enrgy from the battery goes into: chemical bond splitting--is incorrect,as some of the energy is converted to heat by means i have stated above.
When you remove some of that work from the system by way of the moving piston, it has left the building, Elvis is on stage no more and the amount of energy that you can recover from what remains is now diminished.
Incorrect.
1-the work has already been done before the pneumatic cylinder comes into opperation.
2-Elvis has not left the building,the building just got bigger,and no YOU cant find Elvis,so you assume he is missing.
The accounts remain balanced, and your scheme has failed to produce the excess energy you claim.
Nope-my account has grown,and im afraid the only failure is on your behalf Mark.
.You are simply and utterly wrong. Take any volume of gas, and reduce the pressure, IE remove gas, and the remaining energy in the volume has been reduced:
Im begining to think that you make things up as you go Mark-->whats with that?
Please show where we have removed any gas when the piston is opperated?.
NO gas is removed from the system at the point in time when the piston has reached the end of it's travel-->how ever did you come up with-->the volume is reduced when the pressure is reduced.
The pressure is reduced because the storage medium just became larger-->the volume of gas remains the same.
Ideal Gas Law, Charles Law, and Boyle's Law all apply.Yes energy is conserved and no surplus becomes available because of the extra pressure and no extra can be extracted by operating the cylinder. Only by applying funny accounting do you challenge those facts.
Please show us all where these laws state that storing 2ltr's of gas at 100psi has less energy than 2ltr's of gas stored at 10psi. The funny accounting seems to be coming from your way Mark-not mine.
I am not joining you on your stroll down yours.Actually do the math or do the experiment and maybe you will see past your false assumptions.Now you're thinking.
Indeed i am thinking.
Im thinking that you are not quite as smart as we were all led to believe,or you have some motive for steering others astray with your faulse accusations.
And for those exact reasons your proposed scheme fails.Oops, now you've jumped right on your garden path again.Obviously your hand calculator is not working for you. Neither is basic logic.
My logic is sound,but i feel as though it is you that has ran out of fingers Mark.Have you actually read some of the things you have said--even in this post alone.
Lets recap. The below quote was from you
Quote:There you go with the screwy accounting again. The enrgy from the battery goes into: chemical bond splitting, and the phase change.
No-some energy is transformed into heat energy via way of resistive heating.There is definitely some screwy accounting going on,but it is not on my behalf.
Quote:You are simply and utterly wrong. Take any volume of gas, and reduce the pressure, IE remove gas
No where is gas removed in the system when the piston has reached it's maximum travel.
maybe you will see past your false assumptions
There is definitely some faulse assumptions going on here,but there not mine.
Those here can look back over this post Mark,and see some of the incorrect and faulse statements you have made--remember that. I must say that this isnt looking good for you ATM.
I am not joining you on your stroll down yours.
This is probably for the best,as you dont seem to be able to look at everything that is happening within the system. Missing the heat output via way of resistive heating is something that is common knowledge. There is also the leading of others up the garden path on your behalf with your faulse statements,like-Take any volume of gas, and reduce the pressure, IE remove gas.Like i said,no where is gas removed from the system when the pressure is dropped due to the piston in the ram moving to the opposite end of the cylinder.The storage medium has just became larger,but the volume of gas remains-along with its retained energy.
As much as i dont like to say it Mark,but it seems as though your primary goal here is to shut down peoples devices and claims without fully understanding as to what is actually happening throughout the system as a whole.You also assume that i have not done the said and commonly asked for test-this is also another faulse assumption on your behalf.
I think maybe it's time you spent some time on the bench,and catch up a little here with the described setup.
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Lol,this is funny. Please show where the energy was created and destroyed.
LOL Dude, you are the one who claims that by adding a pneumatic ram and letting pressure build up that you can get work for free. The corollary is that if the pressure were reduced below 1 ATM that you would be stuck destroying energy. If you have given up on your preposterous claims then just say so.
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No no no mark E.the environmental heat flow is in contradiction to our gas compression/decompression cycle.it is a gain not a loss
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If you have given up on your preposterous claims then just say so.
I have not given up on anything Mark-->but i feel as though you should maybe try again with your calculations of the energy supplied by the battery and the heat energy conversion(and it's reasons for being produced)being disipated within the circuit. This time,try to include resistive heating along with your phase change heating,and also add to that the heat from the chemical reactions taking place within the battery it self,as the battery will indeed produce heat.
The corollary is that if the pressure were reduced below 1 ATM that you would be stuck destroying energy.
Now is your chance at redemption Mark-->please show us all where the pressure is reduced below 1 ATM within my setup. This is one of those unjustified and incorrect quotes that you often make throughout threads-->this is what people are talking about when they say you are misinterpreting there statements,and adding faulse claims-->things they never said.
LOL Dude, you are the one who claims that by adding a pneumatic ram and letting pressure build up that you can get work for free.
That is the beauty of useing HHO,regardless of wether the pressure is reduced when increasing the size of the storage medium,or is maintained at current pressure with current size storage tank,the energy stored in that HHO gas dose not change.So if we have say a 20 ltr tank,and that is full of HHO with a gage pressure of say 100psi-we would have X amount of energy stored within that HHO gas. Now,if we could expand the size of the tank (without losing any of the stored gas) to say a 30 ltr capacity,our gas pressure would drop,but we would still have the same amount of gas,and the same amount of energy stored within that gas. Feel free to jump on google,and try and find something that can disprove this.
Once you have learned how to account and calculate for all energies,and what gives rise to those energies within my system,please feel free to come back and have another look. But for the time being,it is clear that you are set in your way's,and the stated laws must remain-regardless of wether or not they do. You are a man with a closed mind,and this has become very transparant in this and other threads. Your failure to account for simple things like resistive heating and stored energies in a given amount of gas under different pressures is a sign that your eyes are wide shut-->the laws of physics must stand no matter what-is the attitude you seem to have. This is simply not acceptable in an enviroment such as this forum.
How about you do the testing Mark,and prove i am wrong,as we went by the very laws you praise,and now that i show an extra energy output which is not additive to the supplied energy to the system,you make a back peddle,and start reaching for the stars. The beauty of these forums is that people can go back and read all that was writen. They can go back and see that you agreed with each step(including the storage of gas under pressure (which is on the magnet myths and misconceptions thread),until i showed the extra energy output. Then you went serching for explanations(although incorrect ones)that could explain away what i have shown. But then you started to make mistakes in your calculation,and then started the misdirection's by way of stateing things that are not apparent in my setup-eg,we reduce the amount of gas because we reduced the pressure in the storage tank,and thus reduce the stored energy within that gas. But as we know,we never reduced the amount of gas at all,we simply increased the size of the storage tank.
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No no no mark E.the environmental heat flow is in contradiction to our gas compression/decompression cycle.it is a gain not a loss
These are just some of the things Mark has missed profitis. I suspect there may have been a few other concepts that have been passed on by because of misdirection given in there opperation.
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Start using some engineering calcs guys, talking can only take you so far. The refrigerator cycle not being able to produce useful work was shown to me that way. I kept thinking that 1kw makes 5kw, then use the 5kw to power the 1kw motor. Even with massive losses it could still work! But alas, it was show that it can't.
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Calculations pomodoro: E= RT/nf ln P .whats gona happen to the E if the T goes down a good few kelvin while the p goes up. Whats gona happen if the T goes up while the p comes down.this equasion is clearly fighting with itself in this case.
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I was referring to the complete electrolysis ,combustion process.
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Burning of fuels yes you will lose a lot of heat in the cycle.but the total energy balance remains unaltered : your total in from battery.combustion total out from burning.environment total from phase changes.3 seperate totals.3 seperate entities
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It seems to me the answer to this puzzle is to forget the electrolysis and answer this question: Does the heater produce same heat when supplied by the gas from the tank bypassing the piston, as it does when supplied by the gas that's done the work in the piston. I don't know the answer but on the surface it looks like an easy to solve thermo calc. What you reckon chaps?
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It seems to me the answer to this puzzle is to forget the electrolysis and answer this question: Does the heater produce same heat when supplied by the gas from the tank bypassing the piston, as it does when supplied by the gas that's done the work in the piston. I don't know the answer but on the surface it looks like an easy to solve thermo calc. What you reckon chaps?
This is a pretty simple problem. If the piston does external work, then as I said: that work like Elvis has left the building. Some of the PE in the gas has been sent out of the system in terms of external work and it is no longer available. When the piston merely compresses a spring, then Profitis' best friend the second law comes to the fore.
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Brb
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It seems to me the answer to this puzzle is to forget the electrolysis and answer this question: Does the heater produce same heat when supplied by the gas from the tank bypassing the piston, as it does when supplied by the gas that's done the work in the piston. I don't know the answer but on the surface it looks like an easy to solve thermo calc. What you reckon chaps?
Yes it dose,regardless of what Mark may say. Moving the piston is doing nothing more than enlarging the tank the gas is stored in-->but the amount of gas never changes,thus the energy contained within that gas also never changes-->joules per ltr remains. So regardless of wether the gas go's from the large tank,or via the way of the cylinder first,the same heat energy will be produced by that amount of gas. Mark will give you no reason at all as to how the energy stored within that gas can change just by moving the piston-->because it dosnt change. !ltr of gas is 1ltr of gas regardless of where it must flow to get to the heater. This is why we are useing HHO gas insted of compressed air,as the energy in compressed air is the pressure,where as with HHO,the prssure is a byproduct only-->and it is this byproduct we are gaining our extra energy from.
Mark also states that producing the HHO under pressure requires more energy per mole of gas,and while this is correct,he forgets the fact that more heat energy is produced at the same time-->energy is conserved. So while gas production may go down per joule of energy,heat energy production go's up by the same amount.
Opperating the piston,and gaining energy from this movement comes at no extra cost to the energy input of the system.
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Yes it dose,regardless of what Mark may say. Moving the piston is doing nothing more than enlarging the tank the gas is stored in-->but the amount of gas never changes,thus the energy contained within that gas also never changes-->joules per ltr remains. So regardless of wether the gas go's from the large tank,or via the way of the cylinder first,the same heat energy will be produced by that amount of gas. Mark will give you no reason at all as to how the energy stored within that gas can change just by moving the piston-->because it dosnt change. !ltr of gas is 1ltr of gas regardless of where it must flow to get to the heater. This is why we are useing HHO gas insted of compressed air,as the energy in compressed air is the pressure,where as with HHO,the prssure is a byproduct only-->and it is this byproduct we are gaining our extra energy from.
Mark also states that producing the HHO under pressure requires more energy per mole of gas,and while this is correct,he forgets the fact that more heat energy is produced at the same time-->energy is conserved. So while gas production may go down per joule of energy,heat energy production go's up by the same amount.
Opperating the piston,and gaining energy from this movement comes at no extra cost to the energy input of the system.
Tinman, if the piston performs external work, then the PV product inside the appartus falls: Energy and Elvis have left the system. They aren't coming back. Only you have done your accounting improperly.
If the piston does not perform useful work then the PV product remains constant and the existence of the piston does nothing to change the system. Either way, your premise fails.
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If the compressed gas, say 1L at 120 psi, goes straight to the burner, there will be some cooling as it expands at the nozzle, then the heat from burning. If it goes to the piston instead, assume the cylinder is 1L in volume and valve A is kept open for the whole stroke, then some work is done and we still have 120 psi if the tank is large. Tap B opens, A closes, and the piston pushes the whole 1L to the burner. The heat from the burning will be the same, since 1L at 120 psi was in the cylinder. The pressure will be on average 60 psi, since the cylinder will discharge at 120 initially but nearly zero at the end, when the plunger starts pushing back. There is less cooling at the nozzle as the gas expands since the pressure will average 60 psi for the cycle. However, the gas in the cylinder will cool with expansion, so overall the cooling of the gas in both cases is the same. The pressure in the big tank is lowered slightly by the same amount, but we assumes a very large tank to avoid extra calculations. So yes it seems that you get work out of the gas. No doubt about it.
However there was no need to electrolyse at 120 psi if only burning was required!! In other words, compressing a gas only to burn it later is a waste of energy. The piston system is a way to get some of the work put into compressing it back, since you don't get back it at the flame.
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Tinman, if the piston performs external work, then the PV product inside the appartus falls: Energy and Elvis have left the system. They aren't coming back. Only you have done your accounting improperly.
If the piston does not perform useful work then the PV product remains constant and the existence of the piston does nothing to change the system. Either way, your premise fails.
Once again incorrect.
The PV product dose not fall. P decreases-GV remains the same, and the CA/V increases.
Once again you are missleading people with incorrect information. A liter of water is a liter of water regardless of the size of the bucket it is in.
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Pomodoro.who was taxed in the compression cycle: tinman exclusivey or tinman plus environment.
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Pomodoro.who was taxed in the compression cycle: tinman exclusivey or tinman plus environment.
No idea what it is your trying to say profitis.
As we raise the pressure, less hho gas is produced per joule of energy supplied to the system, but as a result, more heat energy is produced to account for the energy supplied-energy is conserved. As heat is the output we want, the heat is not waste heat, and the energy to raise the pressure in the storage tank has been accounted for.
I have no idea as to how Mark seems to think that lowering the pressure within the storage medium by increasing the storage mediums capacity , lowers the stored energy within the hho gas-as the volume of gas remains the same, and total inch pounds x square inches remains the same.
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Trouble is Tinman Mark doesn't very often get things badly wrong.
Refer to "mathematical analysis of an ideal ZED" started by Mondrasek.
Webby pulled out all the stops with that one, so did Mondrasek.
John.
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Once again incorrect.
The PV product dose not fall. P decreases-GV remains the same, and the CA/V increases.
Once again you are missleading people with incorrect information. A liter of water is a liter of water regardless of the size of the bucket it is in.
Tinman when a gas volume does work in the course of expanding then the P*V product at the end of the stroke is always less than the P*V product at the start of the stroke less the amount of work done.
A liter of water is a liter of (essentially) incompressible liquid and in the context of a heat engine cannot be usefully compared to a compressible fluid such as a mixture of hydrogen and oxygen gasses.
You are free to choose one of the two real possibilities here: 1) That the piston moves without doing external work ( we will make the artificial stipulation that there is no friction loss ), in which case the P*V product does not change, but also no external work is performed by the piston. Or 2) We can perform work with the piston. In that second case, we transfer energy out of the gas into the external load. The P*V product IE the energy in the gas at the end of the stroke is less than the P*V product IE the energy in the gas before the stroke. The difference is ideally equal to the external work performed. In all real cases, the difference is greater than the external work performed.
Go read this short and very accessible article on the subject: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heaeng.html
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Tinman:
I am just going to jump in here as an outsider to give you some perspective. I confess I sucked at thermodynamics and all the gas law business in my chemistry classes. But I think we all intuitively have a sense for PV = nRT. But where I think you are 'missing it' is the concept of working being done on a gas. So you have the ideal gas law, Boyle's las, and Charles' law and other stuff.
I get the sense that you are scoffing at that stuff and playing the 'laws' card. There is always a problem there - the bloody laws are based on common sense, real life, and real life experimentation. You don't get an 'exit card' for that kind of stuff. You are putting up a defensive barrier and seemingingly refusing to accept new ideas. This is very similar to the recent big debate about magnetic fields.
Let me give you my bare-bones simple analogy.
You have a cylinder of air with a piston on one end. The air inside the piston is at one atmosphere and room temperature. Imagine some kind of bench apparatus, a clear cylinder one meter long and 20 cm in diameter with a nice plunger mechanism. You have some gear you can turn by hand to push the piston forward.
Here is what I think you think: You crank the gear my hand and the cylinder moves half-way up the piston. Then you think you have gas that is one-half the volume and the pressure has increased and the temperature of the gas still at room temperature.
But that's not the case!! You cranked the gear and you ended up doing work on the gas. That work has to go somewhere, where did it go? The answer is that not only did the gas pressure go up higher than you think as the volume went down, but the temperature of the gas also went up. The work you did ended up heating up the gas and resulting in 'extra' high pressure and 'extra' temperature. That's the way it works out.
So supposing you now wait for three hours. Well, the hot gas is hotter than the room, so eventually it cools back down to room temperature as air convection currents take the excess heat away. So obviously while the gas was cooling down, the pressure was also lowering because of PV = nRT.
And then finally, what happens if you let the cylinder move back to to the original position? Now the gas is doing work on the outside world. If the gas is exporting energy it comes from the pressure, volume, and the temperature. If the piston actually makes it back to the original starting position, then the gas has to end up colder than the ambient temperature. Kind of like to get the cylinder back to the starting point, the "gas runs out of 'gas'" and since it is pooped out, it has to give up heat energy to get the cylinder to the starting position and therefore the temperature of the gas goes down.
In essence, that's why you see frost on a big air tank that is emptying. If you go to the computer store and you buy a can of compressed air and start to use it, it gets really cold in your hand as you use it.
So just some simple non-technical explanations for what it really means when you compress gas with a piston, or the the gas itself moves a piston.
MileHigh
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Tinman:
Then there is kind of really basic explanation based on simple common sense that ignores all of the laws and top-level energy analysis.
What's a gas? It's like billiard balls bouncing around forever that never lose energy. When two molecules hit they bounce off each other and no energy is lost. Imagine magic ping-pong balls for the air molecules.
So what happens as the cylinder moves forward? The cylinder moving forward is like a giant swinging ping pong bat. As it moves forward, the ping pong balls pick up energy from the bat and start moving faster. Just like hitting a ping pong ball with a bat in real life.
So you end up with the same amount of ping-pong balls in a smaller space, and the ping-pong balls are now more 'energetic' and moving faster. So what's pressure? Pressure is the average number of ping pong balls hitting a small area per second. Obviously the pressure goes up. So what's temperature? Temperature is directly related to molecular motion. Faster and more vigorously moving ping pong balls, the higher the temperature.
Now, if you imagine the cylinder moving backwards. You can think it through. The ping pong ball hitting a wall moving away, bounces back with less kinetic energy after the bounce as compared to before the bounce. So as the 'ping pong bat' moves away, it's almost like it is 'sucking the energy' out of the ping pong balls. Or you can say the ping pong balls are coughing up kinetic energy to make the pistion move backwards. So the pressure goes down and the temperature goes down.
What about when a liquid evaporates? That's simply a liquid ping pong ball getting 'hit out of the park' and taking to the air. There was a mini ping pong bat in the liquid that had to give up energy to 'whack' the ping pong ball out of the park. So as a liquid evaporates it's temperature goes down.
So you can see on a microscopic level that when the piston moves forward, that work goes into the gas. When it moves backward, the gas supplies the work required to move the piston backwards.
MileHigh
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Profitis, who gets taxed? Well, the extra energy required to compress the gas comes from power supply. The piston work comes from there, the power supply. Unless someone has proven that pressure has little effect on the power required from electrolyis. I might be blind but how does the environment give energy to this system?
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Moving the piston is doing nothing more than enlarging the tank the gas is stored in-->but the amount of gas never changes,thus the energy contained within that gas also never changes-->joules per ltr remains.
That is where you go wrong..
the product of P1V1 = P2V2 is only true IF the temperature of the gas remains constant.
Remember, the ideal gas law says p1V1/T1 = p2V2/T2 ... and that's why the subject is called thermo-*dynamics*. There are many ways to expand or compress gas and what happens to the net internal energy of the working fluid depends on the cycle chosen. Your assumptions would only be true for reversible adiabatic processes.
Good luck getting any net work out of such a system, or even building it in the first place.
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I might be blind but how does the environment give energy to this system?
Profitis, who gets taxed? Well, the extra energy required to compress the gas comes from power supply. The piston work comes from there, the power supply. Unless someone has proven that pressure has little effect on the power required from electrolyis.
No-the extra energy drawn from the power supply when the pressure rises is disipated as heat energy due to the resistance dropping between the electrodes in the electrolisis cell. As the resistance has dropped,then the current increases,and as the current has increased,then so dose the heat produced by the cell.
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That is where you go wrong..
the product of P1V1 = P2V2 is only true IF the temperature of the gas remains constant.
Remember, the ideal gas law says p1V1/T1 = p2V2/T2 ... and that's why the subject is called thermo-*dynamics*. There are many ways to expand or compress gas and what happens to the net internal energy of the working fluid depends on the cycle chosen. Your assumptions would only be true for reversible adiabatic processes.
Good luck getting any net work out of such a system, or even building it in the first place.
Here is the problem you have with the !ideal gas law!-HHO is not an ideal gas.
Second thing to remember is that regardless of volume and pressure,the mass !n!(the number of moles) of gas is constant. So regardless of pressure or volume the mass(moles of gas)remain the same. 241.8KJ of energy is produced for every mole of H2 burned.
I repeat--> the mass (moles of gas)remain a constant regardless of pressure or volume
1 mole of H2 will produce 241.8KJ of energy when burned.
Moving the piston with the gas pressure dose not change the mass of the gas-->dose not change the mole amount of gas,and dose not change the stored energy within that fixed mass of gas.
PV is not gas mass(moles of gas) amount.
So as i said,the stored energy within the gas dose NOT change because of a change in PV.
The PV can change without effecting the stored energy within the gas-->the given P can perform work upon the piston without effecting the mass(mole)amount of the gas. As the moles of gas has remained the same,then the energy contained within that gas is unchanged when subject to a pressure drop.
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No-the extra energy drawn from the power supply when the pressure rises is disipated as heat energy due to the resistance dropping between the electrodes in the electrolisis cell. As the resistance has dropped,then the current increases,and as the current has increased,then so dose the heat produced by the cell.
What about the energy used to compress the gas? Potential energy stored in the gas can only come from the battery. Surely the electrolysis stops as the pressure rises, as the Nernst equation demands more voltage due to a higher partial pressure of the gasses? Yes some of the energy goes into heat, the rest into potential energy (pressure) of the gas?
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And the temperature would change why?
The gas temperature will remain at room temperature,as the pressure in the cylinder dose not go any higher that the source that feeds it.
LE's explanation was generic. One can traverse in either direction. In your case, you have a certan amount of battery energy that results in broken chemical bonds and a certain amount of energy in the form of PE in the gas. You then propose to expand the cylinder. Your earliest proposition was that process could deliver external work without removing any of the chemical potential of the gas, which everyone agrees on, but also without losing potential energy from the gas, which multiple people have tried to educate you is flat wrong. As Pomodoro has pointed out you can just set the electrolysis aside as whatever energy goes into breaking the bonds you can get back. Now, all you have is a hot gas engine that you assert can deliver free energy.
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What about the energy used to compress the gas? Potential energy stored in the gas can only come from the battery. Surely the electrolysis stops as the pressure rises, as the Nernst equation demands more voltage due to a higher partial pressure of the gasses? Yes some of the energy goes into heat, the rest into potential energy (pressure) of the gas?
The gas is the stored potential energy,not the pressure at which it is stored. The mass of gas(moles) produced dose not change regardless of pressure,only the volume changes with pressure.When you use your compressor to compress air,the extra load(higher power consumption) on the motor as the pressure builds in the tank is converted into what?-->yes,heat. As we want only compressed air,the heat is seen as waste heat. To raise the pressure of the compressed air,we produce more heat. As this system is designed to produce heat,then the heat energy is not seen as a waste.
So the energy didnt go into pressurising the gas,it went into producing the gas and heat. The heat is the disipated energy that is the pressure.
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Here is the problem you have with the !ideal gas law!-HHO is not an ideal gas.
To the extent that any gas deviates from the ideal gas law it always does so in a lossy way.
Second thing to remember is that regardless of volume and pressure,the mass !n!(the number of moles) of gas is constant. So regardless of pressure or volume the mass(moles of gas)remain the same. 241.8KJ of energy is produced for every mole of H2 burned.
Sure, but the number of moles of H2 that get evolved from the water to later burn for each Joule of battery energy applied shrinks as the vessel pressure increases.
I repeat--> the mass (moles of gas)remain a constant regardless of pressure or volume
In that you are plain wrong.1 mole of H2 will produce 241.8KJ of energy when burned.
Yes, but your assumptiona as to the number of moles evolved is incorrect.Moving the piston with the gas pressure dose not change the mass of the gas-->dose not change the mole amount of gas,and dose not change the stored energy within that fixed mass of gas.
As has been explained to you multiple times now: If the volume expands without doing any external work then the PV product remains fixed. If external work is done, that work is transferred from the gas to the outside world and is forever lost. If you stipulate a constant temperature: then the PV product falls.
PV is not gas mass(moles of gas) amount.
n remains constant. n = PV/(rT).
So as i said,the stored energy within the gas dose NOT change because of a change in PV.
It changes and PV changes when you perform external work with the gas.
The PV can change without effecting the stored energy within the gas
That is at least hypotehtically true.-->the given P can perform work upon the piston without effecting the mass(mole)amount of the gas.
That is also true. As the moles of gas has remained the same,then the energy contained within that gas is unchanged when subject to a pressure drop.
That last bit is fundamentally wrong.
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The gas is the stored potential energy,not the pressure at which it is stored.
Wrong. The product of the number of molecules and their temperature determines the energy which is equal to the pressure * volume product scaled by Boltzmann's constant.The mass of gas(moles) produced dose not change regardless of pressure,only the volume changes with pressure.When you use your compressor to compress air,the extra load(higher power consumption) on the motor as the pressure builds in the tank is converted into what?-->yes,heat. As we want only compressed air,the heat is seen as waste heat. To raise the pressure of the compressed air,we produce more heat. As this system is designed to produce heat,then the heat energy is not seen as a waste.
What you keep missing is that any external work performed by expanding the gas reduces the PV product. IE the gas temperature falls. Now your energy which is proportional to n*T has fallen. It's gone, over, done.
So the energy didnt go into pressurising the gas,it went into producing the gas and heat. The heat is the disipated energy that is the pressure.
That last bit is a tangled mess.
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Quote frm pomodoro: 'I might be blind but how does the environment give
energy to this system?'
Unquote
Where does the heat flow come from in the following conversion liquid>gas.tinman breaks the bonds.environment works for phase change.environment adds work to this system which was not from tinman
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Another analogy : tinman breaks bonds for zinc bromide solution.zinc forms.bromine liquid forms.environment does jakshit contributi.jakshit phase change
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Wrong. The product of the number of molecules and their temperature determines the energy which is equal to the pressure * volume product scaled by Boltzmann's constant. Now your energy which is proportional to n*T has fallen. That last bit is a tangled mess.
What you keep missing is that any external work performed by expanding the gas reduces the PV product.
What you keep missing is that the PV product has nothing to do with the mass amount.
IE the gas temperature falls.
The over all gas temperature remains the same,and has nothing to do with the temperature of the gas coming from the gas injector nozzle insid ethe heater.
It's gone, over, done
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The mass remains,and thus so dose the energy contained within that mass of gas.
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Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
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What you keep missing is that the PV product has nothing to do with the mass amount.
The over all gas temperature remains the same,and has nothing to do with the temperature of the gas coming from the gas injector nozzle insid ethe heater.
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The mass remains,and thus so dose the energy contained within that mass of gas.
Well first dear tinman I have not talked about mass in this discussion. Why you bring it up is a secret of your own. The PV product is a measure of the gas PE, which is also measured by nRT, none of which are mass specific until we talk about particular types of molecules.
What you keep missing is basic thermodynamics that has been established for almost 200 years. If you are just going to proverbially jam your fingers in your ears and shout "na, na, na" there is no discussion.
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I get the impression that Tinman is not recognising that there are two separate and distinct energy components related to the discussion:
1. The chemical energy associated with the H2 and O2 gas resulting from the splitting of the water. Nobody is disputing this.
2. The energy dynamics related to the compression and expansion of the gas itself. It's like the gas is a kind of compressed spring that can store and release energy. In this case it doesn't matter what the chemical composition of the gas is. The discussion about the laws of thermodynamics is all about this aspect.
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Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
Go back and look at my simplistic analogy using the ping pong balls metaphor. Where does the energy come from to evaporate a liquid? How is there a "ping pong bat" in the liquid to launch a ping pong ball out of the park?
That energy comes from the heat stored in the liquid itself or in the ambient environment around the liquid. The nature of the molecular motion is that the molecules have a kind of normal distribution for their velocities. Sometimes there is a high velocity collision in the right direction to launch a ping pong ball out of the park.
There is no energy gain at all. The high velocity ping pong ball that has escaped the liquid is balanced out by the fact that the liquid has lost some thermal energy.
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I get the impression that Tinman is not recognising that there are two separate and distinct energy components related to the discussion:
1. The chemical energy associated with the H2 and O2 gas resulting from the splitting of the water. Nobody is disputing this.
2. The energy dynamics related to the compression and expansion of the gas itself. It's like the gas is a kind of compressed spring that can store and release energy. In this case it doesn't matter what the chemical composition of the gas is. The discussion about the laws of thermodynamics is all about this aspect.
Several people have brought out this distinction. I think that Tinman is fine with it. Tinman also gets that for a set energy condition P*V is a product: Alter V and P follows inversely. But where he gets stuck is that he thinks that P*V product is cast in stone, whether we add or subtract energy from the gas volume. He thinks that the piston can do work without that work coming out of the gas that does the work. It is a very fundamental error.
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Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
If you are going to argue the toss on basic thermodynamics at least study and understand it as it is formulated first.
There are no gains to be had during a phase change. Liberation of energy during a phase change is counterbalanced by the energy requirement for the reverse process to occur. Unless you happen to have continuous supply of a substance undergoing a change (in other words a 'fuel') then the net effect over the cycle is neutral.
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Quote milehigh:' The high velocity ping pong ball
that has escaped the liquid is balanced out by the fact that the
liquid has lost some thermal energy.'
Unquote
Excellent my love.who replaces that lost energy.tinman or the environment
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Quote libre: ' Liberation of
energy during a phase change is counterbalanced by the energy
requirement for the reverse process to occur'
Unquote
If the reversability is more efficient than the environments donated heat flow then we have no problem. Tinman is not really interested in reversability right now.he just wants the accounting balance chekbooks.he wants to know his own total minimum tax in vs total out.you will not be permitted to leave out environmental contribution of work in the phase change liquid>gas and hence compression effort
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@pomodoro: E= RT/nf ln P. When the P rise heats up the tinman device larger than the T drop from phase change..that's when tinmans tax goes bigger than environment's church donations.
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Several people have brought out this distinction. I think that Tinman is fine with it. Tinman also gets that for a set energy condition P*V is a product: Alter V and P follows inversely. But where he gets stuck is that he thinks that P*V product is cast in stone whether we add or subtract energy from the gas volume. He thinks that the piston can do work without that work coming out of the gas that does the work. It is a very fundamental error.
I am not the one stuck here-->lets explain where you are going wrong Mark,MH,and the sheeple that follow.
It is undeniable that when some one here(on this forum) believes they have a device that can deliver more energy than it consume's,the guru'sZ(including my self) demand accurate test measures be made before anything else. Meassurement error is normally the call here.
But when it comes time for the guru's to try and prove some one wrong,near enough seems to be good enough. We will look at each of the below law's,and show why they are not applicable to my setup,and have been missused to explain away a free source of energy.
Boyle's law-->For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
You can express this mathematically as
pV = constant
Is this consistent with pV = nRT ?
You have a fixed mass of gas, so n (the number of moles) is constant.
Now why isnt the above applicable in my system? The answer being that the volume increases when the cylinder is opperated-the pressure decreases when the cylinder is opperated,and thus the temperature decreases when the cylinder is opperated.<--Do not mix up a temperature decrease with a decrease in heat energy within my systemAs the temperature dose not remain constant,boyle's law dose not apply. There is an interesting thing that go's along with boyle's law--Quote:If you want to increase the pressure of a fixed mass of gas without changing the temperature, the only way you can do it is to squeeze it into a smaller volume. That causes the molecules to hit the walls more often, and so the pressure increases. The same applies for decreasing the pressure without changing the temperature of a fixed mass of gas,you just increase the volume to decrease the pressure.
Charles law
For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature.
Now who knows why Charles law dose not apply with my setup?-->the answer is the pressure of the fixed mass of gas drops when the piston is opperated-->it dose NOT remain at a constant pressure.
Quote: Basic thermodynamic processes are defined such that one of the gas properties (P, V, T, or S) is constant throughout the process.
Which is constant within my system when the piston is opperated?.
So you see,Mark has applied the ideal gas law,that is based around an ideal gas(Quote:a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly).
Deviations-the ideal gas law neglects both molecular size and intermolecular attractions, it is most accurate for monatomic gases at high temperatures and low pressures.
So who here can show me that H2O1 is an ideal gas :D --Where H is a fuel and O is an oxidizer for that fuel-->couldnt find a gas further from an ideal gas if we tried.
What is funny to see ,is the sheep in the flock follow Mark blindly without even bothering to do a little reserch them self-->including you MH,the one that insist that i read books to gain knowledge,and yet you failed to do it your self here and now.
@MH,either you didnt read the thread from the start,or you have just blindly followed the deviated and incorrect path that Mark has set.
Fact's
Regardless of any change in temperature.pressure.volume-,the mass of gas remains constant,and the energy stored within that gas remains constant.
When the volume is increased(the piston is opperated)the pressure decreases,and the temperature decreases-->but the heat energy remains the same throughout the system--the heat just dosnt disappear as this would be destroying energy-->the heat energy remains constant regardless of the temperature drop.
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The PV product is a measure of the gas PE, which is also measured by nRT, none of which are mass specific until we talk about particular types of molecules.
Well first dear tinman I have not talked about mass in this discussion. Why you bring it up is a secret of your own.
The whole thread is about the stored energy within the HHO gas-->the mass of gas. You are the one that took it on yourself to deviate from the intended(and well defined) position of this thread.
I have shown and explaind how and where the energy went to increase the pressure of the mass of gas. The energy required to increase the gas mass pressure is accounted for in the form of heat disipated by the system as a whole.
What you keep missing is basic thermodynamics that has been established for almost 200 years. If you are just going to proverbially jam your fingers in your ears and shout "na, na, na" there is no discussion.
As i explained above-and have proven-you have applied laws based around an ideal gas,and also not applicable to my setup. You continually take threads and side track them,and provide faulse and missleading information-->you have done this time and time again.
Now,time for you to back up your claim that these ideal gas laws apply to my system-AS IT IS
Show me the tests carried out by the guru's that developed these 200 year old law's useing HHO gas as there test subject.
Show me a decrease in heat energy within the stored gas when the piston is opperated--you cannot
And the reason you cant is because you know(well you should know) that there is no lose in heat energy within the gas when the piston is opperated-->how is your thermodynamics doing now in this situation.
Next i want you to go and do some reserch on how the ideal gas laws were calibrated-->what type of vessel was used in these experiments. Then come back and tell us all-->i'll give you a little hint-->it was of a fixed volume, Now,is my gas containment vessel of a fixed size when the piston starts to opperate?
About time you put your money where your mouth is Mark,and start showing some truth to your claims.
And a comment from profitis you refuse to challenge because you know he is right.
Quote: Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
Now,spend 5 minutes and go look up profitis's claim,and let us know what you find.
It's time you stopped useing DC power messurements to try and calculate AC power.
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Here's an easy test for this device that anyone can do.the system must be completely sealed no holes.air may be used in place oxygen.column arbitarily long.voltage input must not exceed 0.3-0.5v and calculations must be precise:exact moles of gas = force x distance of column bottom to top
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The whole thread is about the stored energy within the HHO gas-->the mass of gas. You are the one that took it on yourself to deviate from the intended(and well defined) position of this thread.
I have shown and explaind how and where the energy went to increase the pressure of the mass of gas. The energy required to increase the gas mass pressure is accounted for in the form of heat disipated by the system as a whole.
As i explained above-and have proven-you have applied laws based around an ideal gas,and also not applicable to my setup. You continually take threads and side track them,and provide faulse and missleading information-->you have done this time and time again.
Now,time for you to back up your claim that these ideal gas laws apply to my system-AS IT IS
Show me the tests carried out by the guru's that developed these 200 year old law's useing HHO gas as there test subject.
Show me a decrease in heat energy within the stored gas when the piston is opperated--you cannot
And the reason you cant is because you know(well you should know) that there is no lose in heat energy within the gas when the piston is opperated-->how is your thermodynamics doing now in this situation.
Next i want you to go and do some reserch on how the ideal gas laws were calibrated-->what type of vessel was used in these experiments. Then come back and tell us all-->i'll give you a little hint-->it was of a fixed volume, Now,is my gas containment vessel of a fixed size when the piston starts to opperate?
About time you put your money where your mouth is Mark,and start showing some truth to your claims.
And a comment from profitis you refuse to challenge because you know he is right.
Quote: Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
Now,spend 5 minutes and go look up profitis's claim,and let us know what you find.
It's time you stopped useing DC power messurements to try and calculate AC power.
There is probably little point in attempting to educate you in what conventional science has known for > 200 years and what repeated experimental evidence shows.
BUT.. why don't you perform the following experiment. It will show you that your assumptions and theories are completely erroneous.
Take a bike pump and block of the exit with your thumb. Compress the air rapidly... what happens ? Two things should be obvious. You expend energy (work) to move the piston and 2, the air inside the piston heats up. Now, hold the piston in the pumped position till the pump cools down.
Done rapidly enough the first action is approximately equal to and adiabatic process, and over time the heat LEAVES the cylinder (It must do, as it is hotter than the environment).
This experiment alone should be enough to give the lie to your conjecture that the internal energy of a gas is the product of PV and does not change during a thermodynamic process.
Why did the gas heat up? Since internal energy of the gas is proportional to the molecular quantity * temperature and temperature increased then the internal energy of the gas must have increased. Clearly, the work done was converted to heat in the system.
Now, as the gas in the pump cools and heat is transferred to the environment it eventual becomes the same temperature as the environment. The product of PV before compression is the same as after and the internal energy of the gas is the same as before. BUT, at that point all the heat created by the application of work has left the system.
Now, having done this simple experiment go back and analyse your previous statements and realise you are completely misguided about even the simplest thermodynamic processes.
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Tinman,
You are talking "apples" and being answered with "oranges".
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Now, as the gas in the pump cools and heat is transferred to the environment it eventual becomes the same temperature as the environment. The product of PV before compression is the same as after and the internal energy of the gas is the same as before. BUT, at that point all the heat created by the application of work has left the system.
Now, having done this simple experiment go back and analyse your previous statements and realise you are completely misguided about even the simplest thermodynamic processes.
You obviously havnt read the entire thread. Lets see who has it wrong.
BUT.. why don't you perform the following experiment. It will show you that your assumptions and theories are completely erroneous.
We shall see.
Take a bike pump and block of the exit with your thumb. Compress the air rapidly... what happens ? Two things should be obvious. You expend energy (work) to move the piston and 2, the air inside the piston heats up.
At this point you are thinking that it was the work done that caused the air to heat up-right?
Wrong. The work done was only to reduce the volume-the heat already existed within the air. You have taken the heat stored in the air of the whole bike pump,and moved all that heat into a much smaller area-->you just condenced the all ready existing heat.
Backup experiment--> once you have compressed you volume of air,and you now feel the heat build up,you then return the piston of the bike pump back to it's starting position-now the heat has gone-right?-->no,wrong. The heat still exist,and the amount of heat energy wether the bike pump is compressed or uncompressed is exactly the same-->the heat energy amount never changes.
There is probably little point in attempting to educate you in what conventional science has known for > 200 years and what repeated experimental evidence shows.
Time to back up your claim. Please post here on this thread any test carried out on a device that opperates as mine dose-->show us this experimental evidence carried out on a device such as mine,with the same opperating principles.-->My guess is that you will not.
Now, hold the piston in the pumped position till the pump cools down.
And this has to do with what in regards to my system. Im guessing you missed the bit about the whole device being housed in an insulated room that is the same temperature as the gas within the device.
Done rapidly enough the first action is approximately equal to and adiabatic process, and over time the heat LEAVES the cylinder (It must do, as it is hotter than the environment).
You mean that it leaves the cylinder to be disipated into a room at the same temperature ???-->as in the case of my device.
This experiment alone should be enough to give the lie to your conjecture that the internal energy of a gas is the product of PV and does not change during a thermodynamic process.
You are doing a MarkE. I have stated that the energy stored within the HHO gas mass dose not change regardless of the PV product,and this is what the thread was about--im guessing you missed that bit to :o
Why did the gas heat up? Since internal energy of the gas is proportional to the molecular quantity * temperature and temperature increased then the internal energy of the gas must have increased. Clearly, the work done was converted to heat in the system.
The highlighted i have already explained--it is because you condenced the already avaliable heat within the cylinder.
No,the internal heat energy did not increase,as it was already there--explained twice already.
And no-the work done was to compress the gas,the heat already existed.
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Tinman,
You are talking "apples" and being answered with "oranges".
Yes camelherder-->this is what the guru's do,although there are a couple that are realistic here on the forum.
They are using a 200 year old formula that do not suit the system i have show. Either the P,V or T must be constant,and in my system none of them are,they are all veriables. Also the ideal gas laws(now 200 years old) are based around an ideal gas(a hypothetical ideal gas),and H2O1 is far from an ideal gas--but they continue to peddle there ways.
I am happy that you spotted this camelherder.
Cheers.
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There is probably little point in attempting to educate you in what conventional science has known for > 200 years and what repeated experimental evidence shows.
BUT.. why don't you perform the following experiment. It will show you that your assumptions and theories are completely erroneous.
Take a bike pump and block of the exit with your thumb. Compress the air rapidly... what happens ? Two things should be obvious. You expend energy (work) to move the piston and 2, the air inside the piston heats up. Now, hold the piston in the pumped position till the pump cools down.
Done rapidly enough the first action is approximately equal to and adiabatic process, and over time the heat LEAVES the cylinder (It must do, as it is hotter than the environment).
This experiment alone should be enough to give the lie to your conjecture that the internal energy of a gas is the product of PV and does not change during a thermodynamic process.
Why did the gas heat up? Since internal energy of the gas is proportional to the molecular quantity * temperature and temperature increased then the internal energy of the gas must have increased. Clearly, the work done was converted to heat in the system.
Now, as the gas in the pump cools and heat is transferred to the environment it eventual becomes the same temperature as the environment. The product of PV before compression is the same as after and the internal energy of the gas is the same as before. BUT, at that point all the heat created by the application of work has left the system.
Now, having done this simple experiment go back and analyse your previous statements and realise you are completely misguided about even the simplest thermodynamic processes.
Here is some information for you libre-your own Boyle's law states-
Quote:At the molecular level, the pressure of a gas depends on the number of collisions its molecules have with the walls of the container. If the pressure on the piston is doubled, the volume of the gas decreases by one-half. The gas molecules, now confined in a smaller volume, collide with the walls of the container twice as often and their pressure once again equals that of the piston.
How does Boyle's Law relate to the kinetic molecular theory? The first postulate of the theory states that a gas sample occupies a relatively enormous empty space containing molecules of negligible volume. Changing the pressure on the sample changes only the volume of that empty space - not the volume of the molecules.
So did the work done on the piston create the heat?,or was the work done to compress the already existing heat into a smaller area?
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Libre says:'
Take a bike pump and block of the exit with your thumb.
Compress the air rapidly... what happens ? Two things should be
obvious. You expend energy (work) to move the piston and 2, the
air inside the piston heats up.'
The environment contributes to expansion of gas in our case: h20 plus electricity> h2 + o2.unless you can explain to the panel of establishment physicists who are watching this thread how it is exclusively tinman's work.
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I am not the one stuck here-->lets explain where you are going wrong Mark,MH,and the sheeple that follow.
It is undeniable that when some one here(on this forum) believes they have a device that can deliver more energy than it consume's,the guru'sZ(including my self) demand accurate test measures be made before anything else. Meassurement error is normally the call here.
But when it comes time for the guru's to try and prove some one wrong,near enough seems to be good enough. We will look at each of the below law's,and show why they are not applicable to my setup,and have been missused to explain away a free source of energy.
Boyle's law-->For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure.That means that, for example, if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
You can express this mathematically as
pV = constant
Is this consistent with pV = nRT ?
You have a fixed mass of gas, so n (the number of moles) is constant.
Boyle's Law, like the ideal gas law, and Charles' Law, applies to static conditions.
Now why isnt the above applicable in my system? The answer being that the volume increases when the cylinder is opperated-the pressure decreases when the cylinder is opperated,and thus the temperature decreases when the cylinder is opperated.<--Do not mix up a temperature decrease with a decrease in heat energy within my systemAs the temperature dose not remain constant,boyle's law dose not apply. There is an interesting thing that go's along with boyle's law--Quote:If you want to increase the pressure of a fixed mass of gas without changing the temperature, the only way you can do it is to squeeze it into a smaller volume. That causes the molecules to hit the walls more often, and so the pressure increases. The same applies for decreasing the pressure without changing the temperature of a fixed mass of gas,you just increase the volume to decrease the pressure.
Charles law
For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature.
Now who knows why Charles law dose not apply with my setup?-->the answer is the pressure of the fixed mass of gas drops when the piston is opperated-->it dose NOT remain at a constant pressure.
Quote: Basic thermodynamic processes are defined such that one of the gas properties (P, V, T, or S) is constant throughout the process.
Which is constant within my system when the piston is opperated?.
So you see,Mark has applied the ideal gas law,that is based around an ideal gas(Quote:a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly).
Deviations-the ideal gas law neglects both molecular size and intermolecular attractions, it is most accurate for monatomic gases at high temperatures and low pressures.
So who here can show me that H2O1 is an ideal gas :D --Where H is a fuel and O is an oxidizer for that fuel-->couldnt find a gas further from an ideal gas if we tried.
What is funny to see ,is the sheep in the flock follow Mark blindly without even bothering to do a little reserch them self-->including you MH,the one that insist that i read books to gain knowledge,and yet you failed to do it your self here and now.
@MH,either you didnt read the thread from the start,or you have just blindly followed the deviated and incorrect path that Mark has set.
Fact's
Regardless of any change in temperature.pressure.volume-,the mass of gas remains constant,and the energy stored within that gas remains constant.
When the volume is increased(the piston is opperated)the pressure decreases,and the temperature decreases-->but the heat energy remains the same throughout the system--the heat just dosnt disappear as this would be destroying energy-->the heat energy remains constant regardless of the temperature drop.
You have latched onto a little piece of knowledge: gas laws under static conditions and misapplied them to dynamic as in thermodynamic situations. We can transfer thermal energy in and out of a gas volume, and that changes the temperature. When we change the temperature, then for a given volume and number of molecules of gas, the pressure changes accordingly. Pressure against a surface gives us force, and force applied through a distance gives us work, and whoopee, now we can build machines that convert heat to mechanical work by moving that heat through gas volumes. As we impart more energy to a given gas volume the temperature rises, as we remove energy from a given gas volume the temperature falls.
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The whole thread is about the stored energy within the HHO gas-->the mass of gas. You are the one that took it on yourself to deviate from the intended(and well defined) position of this thread.
You have latched onto Charles' Law which is a special case of the Ideal Gas Law. I made no such restriction. Moles in "n" always gives the right answer. Under the special conditions that we have a particular gas, a mole of that gas has a fixed mass, and we can use mass as a proxy for "n" for that gas. Special cases do not define the general case.
I have shown and explaind how and where the energy went to increase the pressure of the mass of gas. The energy required to increase the gas mass pressure is accounted for in the form of heat disipated by the system as a whole.
Choose which of the two real circumstances you want to work with: The cylinder performs external work, or it doesn't. In either case your premise fails.As i explained above-and have proven-you have applied laws based around an ideal gas,and also not applicable to my setup. You continually take threads and side track them,and provide faulse and missleading information-->you have done this time and time again.
I spend the time to try and explain things to you out of respect for the fact that you actually perform experiments to try and find the truth. However, when no amount of patient explanation seems able to dissuade you from your misconceptions we are going to reach a point where it is just a complete waste of time.
You are free to judge for yourself and must judge for yourself whether what anyone says is: correct, mistaken, or intentionally wrong. On these boards there are plenty of examples of each. When doing so, you might wish to compare what you are told by any person or any group of people against other references that have good reputations, and/or show proofs.
Now,time for you to back up your claim that these ideal gas laws apply to my system-AS IT IS
Others and myself have already done that.Show me the tests carried out by the guru's that developed these 200 year old law's useing HHO gas as there test subject.
Start with Carnot.Show me a decrease in heat energy within the stored gas when the piston is opperated--you cannot
Again start with Carnot.And the reason you cant is because you know(well you should know) that there is no lose in heat energy within the gas when the piston is opperated-->how is your thermodynamics doing now in this situation.
They are doing just great. Avail yourself to another teacher because I have been ineffective with you. This may help you: https://www.khanacademy.org/science/physics/thermodynamics/v/work-from-expansionNext i want you to go and do some reserch on how the ideal gas laws were calibrated-->what type of vessel was used in these experiments. Then come back and tell us all-->i'll give you a little hint-->it was of a fixed volume, Now,is my gas containment vessel of a fixed size when the piston starts to opperate?
About time you put your money where your mouth is Mark,and start showing some truth to your claims.
And a comment from profitis you refuse to challenge because you know he is right.
Quote: Mark E and libre will continue to ignore the most important aspect of the system @tinman.the phase change liquid>gas.the gain is here and here alone
Now,spend 5 minutes and go look up profitis's claim,and let us know what you find.
It's time you stopped useing DC power messurements to try and calculate AC power.
If you are concerned with going down garden paths, then you should be very wary of what Profitis writes.
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Remove the burner and electrolysis, they make no difference. Now prove the gas in the pipe from the piston, still at 120 psi, is cooler from doing work, and there is your proof. Entropy is the key here as there is no phase change. Try calcs with piston with no load , then piston with full load. All I know is that when I transfer argon at 4500 psi from onebottle to a smaller one thats empty, one gets cool, but the other gets very warm.
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Pomodoro says:'Remove the burner and electrolysis, they make no difference.'
I say: E= RT/nf ln P.you have no choice but to trust this equasion.it was given to us by the establishment textbooks.in other words it is impossible to ignore the cancelling effect that a change in T has on E.
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mark E says:'be very wary of what Profitis writes.'
I say:' you will be utterly and helplessly forced by established physics to pay attention to the effect of temperature change on the most powerful equasion given to us by school textbooks: the nernst E= RT/nf ln P'
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So did the work done on the piston create the heat?,or was the work done to compress the already existing heat into a smaller area?
Have a look at http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heat.html and answer the question yourself...
In case you are too lazy to do that, lets start with an easily verified fact.
If you compress a gas (ideal or otherwise) the temperature rises. (lets assume an insulated cylinder and rapid compression giving approximately adiabatic conditions.. ) You would agree?
If the amount of substance did not change then we know for sure the internal energy of gas has increased since we both agree that internal energy is proportional to the quantity of stuff * temperature only.
Work and heat are interchangeable concepts.. one is converted to the other and vice versa during thermodynamic processes.
If you don't accept that then there is no hope for you ever understanding thermodynamics or realising why the process you propose does not work as you think.
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Tinman:
I didn't read your thread, I was only chiming in on the thermodynamics stuff.
Your concept about the amount of heat in the gas remaining the same, and instead it just gets hotter when the volume is smaller, is flawed.
Let's take the same number of moles of gas in two different sized containers. Let's say the temperature is the same for both gasses. So we know that P1V1 = P2V2. So if the second container is 1/2 the size the pressure is double.
How much heat is in each container under these conditions? Well the mass is the same and the temperature is the same. You have to measure the amount of heat relative to another temperature, so let's use absolute zero. You have the same type of gas, and you have the same mass. So for both cases you have the same amount of heat energy in the gas, by mass. We are talking about the heat energy that you could extract from the mass of gas in each container. In other words it's like asking now much ice each container could melt if you put each container in a bucket of ice. It would be the same. So your argument about the temperature going up because the same amount of heat has to go into a smaller container is false. Both containers contain the same amount of heat and they are at the same temperature. And that's a big clue right there.
Here is a good comparison: Imagine instead of gas, we use a piece of sponge. So we do the test in a vacuum. The sponge gets compressed down to 1/2 size to fit in the smaller container. Did the temperature of the sponge go up? The answer is no. Did the amount of heat in the sponge go up? The answer is no. So you compressed the heat of the sponge into a smaller volume and the amount of heat stayed the same and the temperature stayed the same. The energy went somewhere else, the sponge also acts like a spring and the energy is stored in the compressed spring.
But in a gas there is no mechanical equivalent of a spring. The only way the gas can store the work done on it by the piston is by having he molecules move faster and store the piston energy as kinetic energy. Higher kinetic energy equals higher temperature.
Let's so back to the one-meter transparent cylinder with the plunger. If you compress the piston very rapidly and then let go, the piston will return back to the start position.
So what happens if you compress the piston and then wait three hours for the compressed gas to cool down and then you release the piston?
The answer is that the piston will not make it back to the start position. It will push back right away when you release it and then say travel 3/4 of the way to the start position. Then it will almost stop dead. Then as you watch it, it will slowly creep back over time to get back to the start position. It might take 10 or 15 minutes before the piston finally makes it back to the start position.
MileHigh
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Tinman:
I didn't read your thread, I was only chiming in on the thermodynamics stuff.
Your concept about the amount of heat in the gas remaining the same, and instead it just gets hotter when the volume is smaller, is flawed.
Let's take the same number of moles of gas in two different sized containers. Let's say the temperature is the same for both gasses. So we know that P1V1 = P2V2. So if the second container is 1/2 the size the pressure is double.
How much heat is in each container under these conditions? Well the mass is the same and the temperature is the same. You have to measure the amount of heat relative to another temperature, so let's use absolute zero. You have the same type of gas, and you have the same mass. So for both cases you have the same amount of heat energy in the gas, by mass. We are talking about the heat energy that you could extract from the mass of gas in each container. In other words it's like asking now much ice each container could melt if you put each container in a bucket of ice. It would be the same. So your argument about the temperature going up because the same amount of heat has to go into a smaller container is false. Both containers contain the same amount of heat and they are at the same temperature. And that's a big clue right there.
Here is a good comparison: Imagine instead of gas, we use a piece of sponge. So we do the test in a vacuum. The sponge gets compressed down to 1/2 size to fit in the smaller container. Did the temperature of the sponge go up? The answer is no. Did the amount of heat in the sponge go up? The answer is no. So you compressed the heat of the sponge into a smaller volume and the amount of heat stayed the same and the temperature stayed the same. The energy went somewhere else, the sponge also acts like a spring and the energy is stored in the compressed spring.
But in a gas there is no mechanical equivalent of a spring. The only way the gas can store the work done on it by the piston is by having he molecules move faster and store the piston energy as kinetic energy. Higher kinetic energy equals higher temperature.
Let's so back to the one-meter transparent cylinder with the plunger. If you compress the piston very rapidly and then let go, the piston will return back to the start position.
So what happens if you compress the piston and then wait three hours for the compressed gas to cool down and then you release the piston?
The answer is that the piston will not make it back to the start position. It will push back right away when you release it and then say travel 3/4 of the way to the start position. Then it will almost stop dead. Then as you watch it, it will slowly creep back over time to get back to the start position. It might take 10 or 15 minutes before the piston finally makes it back to the start position.
MileHigh
It is your logic that is flawed MH, as you just disipated the heat energy into the enviroment. Push the piston in, and the heat is confined into a smaller volume-then let the piston go straight away, and it will return to the starting position.
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@MH&MarkE
If we are to use your ideal gas laws, then in all fairness we are to use an ideal piston/cylinder for this experiment.
The pistion is in the start position-there is no friction in this cylinder/pistion setup. Put a pressure of 5psi gage pressure into the cylender. The cylinder has an ideal insulation. Now push the piston 3/4 of the way down the cylender. The pressure rises, and the temp rises. Leave in the compressed state for 1minute, then let the piston return to the starting point. The pressure will once again be 5psi-all energy has been returned and yet there was a temperature increase.
What caused the temperature increase if all the energy was returned?
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It is your logic that is flawed MH, as you just disipated the heat energy into the enviroment. Push the piston in, and the heat is confined into a smaller volume-then let the piston go straight away, and it will return to the starting position.
This is only true for an adiabatic process, where no thermal energy enters or leaves the system.
Push the piston in and it heats up. Work is converted to heat. The temperature in the system is raised and P1V1 != P2V2.
If you then held the volume constant and allow all the extra heat that was converted from work to transfer back to the environment THEN P1V1 == P2V2 and the internal energy will also be same as the starting condition.
I don't know why you don't seem to be able to grasp such a simple concept, but by refusing to accept it is shows any other conclusions you have come to about your system to be false.
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@MH&MarkE
If we are to use your ideal gas laws, then in all fairness we are to use an ideal piston/cylinder for this experiment.
Sure, I offered that we ignore friction loss before. That leaves two possibilities: The piston movement performs external work, removing more than as much energy in the process, or it does not external work. Either way your premise fails.The pistion is in the start position-there is no friction in this cylinder/pistion setup. Put a pressure of 5psi gage pressure into the cylender. The cylinder has an ideal insulation. Now push the piston 3/4 of the way down the cylender. The pressure rises, and the temp rises. Leave in the compressed state for 1minute, then let the piston return to the starting point. The pressure will once again be 5psi-all energy has been returned and yet there was a temperature increase.
What caused the temperature increase if all the energy was returned?
Well if your cylinder is insulated, then your description is not realizable in this world, friction or no friction in the cylinder. You input energy into the system, and that energy shows up as a higher T in n*R*T. Unsurprisingly, it also shows up as a higher P*V product. When you let the piston move out unopposed to its original position, then to your great surprise you would find that the pressure is greater than the 5psi gage that you had at the start.
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Milehigh says:'So what happens if you compress the piston and then wait three
hours for the compressed gas to cool down and then you release
the piston?'
I say: WHO compressed the piston.it was two individuals who dumped work into the system.two were taxed.tinman is piggybacking ontop of the other guys efforts and gaining in the process.the other guy is called mr environment.mr environment is dumping heat into the begin of compression cycle.mr environment is also reclaiming his heat at end of cycle.
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@MH&MarkE
If we are to use your ideal gas laws, then in all fairness we are to use an ideal piston/cylinder for this experiment.
The pistion is in the start position-there is no friction in this cylinder/pistion setup. Put a pressure of 5psi gage pressure into the cylender. The cylinder has an ideal insulation. Now push the piston 3/4 of the way down the cylender. The pressure rises, and the temp rises. Leave in the compressed state for 1minute, then let the piston return to the starting point. The pressure will once again be 5psi-all energy has been returned and yet there was a temperature increase.
What caused the temperature increase if all the energy was returned?
So why don't perform this experiment. If you do you'll realise you have been mistaken all along.
What you will find is that when compressed the temperature of the gas rises. The rise in proportional to the amount of work done to compress it. The WORK is converted to HEAT...
When it decompresses the HEAT is converted to WORK. It is as simple as that.
What happens in reality of course is that we don't have a perfectly insulated cylinder and heat losses occur. However these losses
don't ever add to the efficiency of the conversion process. Likewise the fact that our working fluid is not a perfect gas.
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Mr libre says:'What you will find is that when compressed the temperature of
the gas rises.'
I say: minus the cooling effect of phase-change liquid>gas
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Mr libre says:'What you will find is that when compressed the temperature of
the gas rises.'
I say: minus the cooling effect of phase-change liquid>gas
And then perhaps you may have re-invented the refrigerator.
Hopefully you are not going to describe that as a free energy device...
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Libre says: 'a refrigerator- hopefully you are not going to describe that as a free energy
device...'
I say: of course not don't be silly.tinmans device is a free energy device
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This is only true for an adiabatic process, where no thermal energy enters or leaves the system.
Push the piston in and it heats up. Work is converted to heat. The temperature in the system is raised and P1V1 != P2V2.
If you then held the volume constant and allow all the extra heat that was converted from work to transfer back to the environment THEN P1V1 == P2V2 and the internal energy will also be same as the starting condition.
I don't know why you don't seem to be able to grasp such a simple concept, but by refusing to accept it is shows any other conclusions you have come to about your system to be false.
Do you have a mind of your own? Or do you just repeat what others have already said-and whats with all the googlegasms you keep having?.
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So why don't perform this experiment. If you do you'll realise you have been mistaken all along.
What you will find is that when compressed the temperature of the gas rises. The rise in proportional to the amount of work done to compress it. The WORK is converted to HEAT...
When it decompresses the HEAT is converted to WORK. It is as simple as that.
What happens in reality of course is that we don't have a perfectly insulated cylinder and heat losses occur. However these losses
don't ever add to the efficiency of the conversion process. Likewise the fact that our working fluid is not a perfect gas.
Nope-the heat already exist, the work done is to reduce the volume of the gas, and thus concentrate the heat . No work is done to create heat.
And you think I havnt done any experimental work toward this? Lol..
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Nope-the heat already exist, the work done is to reduce the volume of the gas, and thus concentrate the heat . No work is done to create heat.
And you think I havnt done any experimental work toward this? Lol..
You really need to do some studying.
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You really need to do some studying.
You really need to get out of the past.-200 year old calculations based on a gas that dosnt exist-thats the ticket :D
Now in regards to my setup, what energy do you think Elvis took with him when he left the building?
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Nope-the heat already exist, the work done is to reduce the volume of the gas, and thus concentrate the heat . No work is done to create heat.
And you think I havnt done any experimental work toward this? Lol..
Absolute BS.
When you push on the cylinder the gas is moved, does it not?
So what is heat? It is the simply the sum of all the kinetic energy of the molecules. Since we 'moved' the molecules using an external force the kinetic energy of the molecules must have increased. Therefore the heat content of the gas rises. It is NOT due to 'existing heat' somehow being concentrated.
If not, where did the work we expended go to? Free space? the moon?, exited via a black hole? The only place it goes is into the heat content of the gas.
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Do you have a mind of your own? Or do you just repeat what others have already said-and whats with all the googlegasms you keep having?.
googlegasms ? Well I admit to having refreshed my thermodynamics knowledge courtesy of the internet on multiple occasions recently but I can still remember learning this as part of an engineering degree some 30 years ago. We used the knowledge modelling heat engines on software and then actually building, instrumenting and testing them.
The agreement of the theory to practical experiment was as accurate then as it is today.
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You really need to get out of the past.-200 year old calculations based on a gas that dosnt exist-thats the ticket :D
Now in regards to my setup, what energy do you think Elvis took with him when he left the building?
Tinman I really do not know what has gotten into you. You have become obstinately ignorant and shrill. As have several others: I've explained to you repeatedly what would occur in your set-up, and pointed you at suitable references. If you choose to be deliberately ignorant so be it. If you choose to act like a jerk at the same time, then again so be it.
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Tinman you have made some points that have no backing or logic to them. The example that comes to mind is the false argument that "HHO" was not like an ideal gas and therefore what Mark was talking about didn't apply. For some reason you were suggesting that the ideal gas laws applied to a different gas, or a gas consisting or only one element, and that was not comparable to "HHO."
Why would you say that? Hydrogen is a gas like any other gas. Oxygen the same. Both follow the "bouncing balls" analogy to the letter. A mixture of hydrogen and oxygen gas will undergo thermodynamic compression/expansion cycles just like any other gas and behave like any other gas. These gasses are most likely operating under conditions where the ideal gas law relationships are perfectly linear. It simply makes no sense to advance the argument that "HHO" is different from H2 of O2 or N2 or just plain air because that is obviously not the case. I am assuming that your experiment does not push the gasses into any extreme regions where the ideal gas law deviates from linearity. But even if it did, the "HHO" gas would still behave like any other gas.
MileHigh
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MH Tinman has gotten himself really, really lost. He'll find out sooner or later just how badly he has deceived himself when he tries to make something based on his misconceptions work.
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I don't think so.
First there is the question that you have NOT answered,, by compressing the gas into a smaller volume does that actually add heat energy or only raise the temperature, there is a difference between the heat energy stored and the temperature of the gas,, basic stuff there so there should be a basic answer,, or maybe all those before you were just really really lost.
The internal energy of the gas is the product: n (number of molecules) * R (Boltzmann's constant) * T (absolute temperature in Kelvin). That means if you've got a certain amount of gas, the energy of that gas is proportional to the absolute temperature. Under static conditions, the product P*V is equal to the internal energy of the gas.
You seem to be forgetting that the temperature of the gas will go up with the decrease in volume and that increase goes along with the increase in pressure,, see work in changes the potential of the system but NOT the stored heat energy, and that energy value is determined at the start and is therefore a static condition that is then set in motion to make a full cycle and returns to the same static condition.
That tangled text really makes no sense in our physical world. n*R*T is an absolute measure of the internal energy of an ideal gas. Real gasses come very close to the ideal. Add energy as heat by any means, or add energy by performing mechanical work and n, and R being constant, T being the only variable, rises. The work increases: both the pressure and the P*V product.
After all Mark,, if this were either at absolute zero Pa or K it would be at both,, they are a balance,, simple really.
Simple it indeed is. Unfortunately, you still managed to get it wrong. The hang-up that you seem to share with Tinman is that you do not recognize that it takes an exchange of work to change the pressure or volume of a fixed amount of gas. So, while it is perfectly feasible to store 1J of energy at say 20C in an infinite number of different pressure and volume combinations, changing from one combination to another with higher pressure requires work that then results in a greater P*V product (bicycle pump gets hot), or to another with a lower pressure releases work that results in a lower P*V product (can of dusting spray gets cold).
An adiabatic process would change the pressure / volume relationship without exchanging any energy with the outside world. That would keep the energy inside Tinman's system constant, no external work would be performed and his premise of performing outside work would fail. A real process exchanges work with the outside world, specifically, the piston is supposed to do something useful when it moves, the energy to do that cools the gas, just like a can of dusting spray cools when gas is released, and the n*R*T has now gone down inside Tinman's system, meaning that it has given up the energy that performed the external work: Elvis has left the building. And again his premise fails.
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MH Tinman has gotten himself really, really lost. He'll find out sooner or later just how badly he has deceived himself when he tries to make something based on his misconceptions work.
Im lost? lol.
Quote MarkE-When you let the piston move out unopposed to its original position, then to your great surprise you would find that the pressure is greater than the 5psi gage that you had at the start.
Wow Mark,you have just invented an OU device. We now have more stored force than we started with. How did you do that when it was clearly stated that it was a frictionless device.
Tinman I really do not know what has gotten into you. You have become obstinately ignorant and shrill.
Well if that isnt the pot calling the kettle black.
You shun my work simply because it dosnt comply with your 200 year old physics. If we dont all agree with you,then we become shrills--.now that is being ignorant. By the way,many of us are still waiting for your physics to tell us what the magnetic force is,as your physics knows all about magnetic fields!right?!
If you choose to act like a jerk at the same time, then again so be it.
A jerk would be some one that just believes what they read,and that is the be all and end all-->not having done any experimental work at all on the system in question.
I've explained to you repeatedly what would occur in your set-up
im sorry,but !Elvis has left the building! is just not enough. Tell me exactly what energies you think have been depleted(or transfered outside the system) in my system when the piston is opperated.
I will then show you that Elvis is dead,and simply cannot leave the building.
It is funny as an engineer watching you physics guys meassureing small parts of a system after an action within that system has taken place,while over looking everything that took place over the system as a whole after that action. You deal in ideals that dont exist,and messurements are made from these ideals. Then you have variables to make things work out when things dont quite add up. As an engineer,we deal in absolutes,and systems as a whole.
Just found this
A real process exchanges work with the outside world, specifically, the piston is supposed to do something useful when it moves, the energy to do that cools the gas, just like a can of dusting spray cools when gas is released, and the n*R*T has now gone down inside Tinman's system, meaning that it has given up the energy that performed the external work:
This is a clasic example of what i was saying in the above statement-failing to look at the complete process.-->The total heat energy is still in the system as a whole-->no heat energy has left the system-Elvis is dead,and cannot walk. This is where your 200 year old physics,and your own blindness Mark,MH and the new guru libre are going to come unstuck. This is the very same attitude that holds all man kind back from being free of fossil fuels. 365 days in a year-oh but every 4 years we have to throw another day in there to make things all work out--screwed up with out time definitions there some where-->but hey,what the hell,it's all good.
Now,which of you knows about/or the mathematics to equate total force within a sealed vessel?-->you know,force per unit area x area total. And which of you is able to calculate total heat energy within a given volume? Lets put engineering up against your physics,and we'll see where that lead's us.
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googlegasms ? . We used the knowledge modelling heat engines on software and then actually building, instrumenting and testing them.
The agreement of the theory to practical experiment was as accurate then as it is today.
but I can still remember learning this as part of an engineering degree some 30 years ago
Well it's time you went back to engineering school,as things have changed. We now look at systems as a whole,and all events that will take place around that system in question. Foe example,when meassureing pressure in a sealed vessel,it is no longer acceptable to use psi. You now have to account for total force over a given(total) area.
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Here is a diagram of the direction of environmental heat flow contribution to and from the tinman device derived exclusively from phase-change liquid>gas gas>liquid
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Mark,, don't take this the wrong way,, but,, I get a kick out of it when someone smarter than me is making the same mistake as I was.
We all make mistakes. But I assure you there is no mistake here.
You are confusing "heat" with "temperature".
Not at all. Heat is energy. (More specifically it is thermal energy in the process of moving from a higher to a lower temperature. In most discussions where the term "heat" is used we are really talking about the internal energy: nRT.) Temperature is a measure of the intensity of heat energy: Its ability to convey energy to its surroundings. The faster the motion of molecules in a gas the easier it is for them to transfer energy to their surroundings, the hotter they are, and the higher their temperature. In order to get the quantity of energy we usually call heat we need to know how many molecules we have: the "n" in nRT.
Heat,, in a very simple way of looking at it, is the rate of motion.
No, that is temperature. The rate of motion is proportional to how easily a mass can give off energy. We have to know how much "stuff" is moving at a given average speed to know how much energy there is. The grade school explanation compares a red hot nail to a bath tub of lukewarm water. The red hot nail has a high temperature. The bath tub has many times more internal (heat) energy.
Temperature,, in a very simple way of looking at it, is the rate of collisions.
See above.
I suppose that Feynman could be wrong on that.
No, he isn't. You should reread him.
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Tinman:
I have been doing this stuff in my head mostly, trying to rely on my wits. I have looked at formulas from time to time but it's mostly from thinking through the problem.
Going back to the transparent cylinder with the piston, I said that if you let the heat escape from the compressed gas first, and then let the piston return by itself, it would not return all the way back in one stroke. It would slowly creep back last 1/4 of the way. We are assuming the plunger is frictionless, but otherwise it's a real bench setup.
Do you agree with that? If yes, when the piston stops after moving 3/4 of the way back to the start position, what is the temperature of the gas?
MileHigh
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Quote from: MarkE on Today at 01:09:25 PM
MH Tinman has gotten himself really, really lost. He'll find out sooner or later just how badly he has deceived himself when he tries to make something based on his misconceptions work.
Im lost? lol.
Yes, you are very badly lost on very fundamental issues.
Quote MarkE-When you let the piston move out unopposed to its original position, then to your great surprise you would find that the pressure is greater than the 5psi gage that you had at the start.
Wow Mark,you have just invented an OU device. We now have more stored force than we started with. How did you do that when it was clearly stated that it was a frictionless device.
No, this is again where you are just lost. Please avail yourself to any of the many thousands of reputable online references because you stopped paying attention to any of the correct explanations offered by others and myself here a long time ago.
Quote
Tinman I really do not know what has gotten into you. You have become obstinately ignorant and shrill.
Well if that isnt the pot calling the kettle black.
You shun my work simply because it dosnt comply with your 200 year old physics. If we dont all agree with you,then we become shrills--.now that is being ignorant. By the way,many of us are still waiting for your physics to tell us what the magnetic force is,as your physics knows all about magnetic fields!right?!
You have made a series of incorrect statements that are refuted left and right by that "200 year old" physics that you disdain. Experiment after experiment refutes your assertions.
Quote
If you choose to act like a jerk at the same time, then again so be it.
A jerk would be some one that just believes what they read,and that is the be all and end all-->not having done any experimental work at all on the system in question.
My definition is one who engages in personal attack rather than debate the merits of their arguments. "Yes it is!" and "No it isn't!" are declarations, not arguments.
Quote
I've explained to you repeatedly what would occur in your set-up
im sorry,but !Elvis has left the building! is just not enough. Tell me exactly what energies you think have been depleted(or transfered outside the system) in my system when the piston is opperated.
I will then show you that Elvis is dead,and simply cannot leave the building.
For the umpteenth time, if the piston does work on the outside world, then it transfers energy to the outside world. The gas exerts force on the piston. It exerts that force in the direction of motion. It performs work. That work comes from the internal energy of the gas. Going the opposite way: External force must be applied to the piston in order to force the gas into a smaller volume. That force is again applied in the direction of motion. That is external work that is performed on the gas. That work increases the internal energy of the gas. It increases nRT and consequently the PV product just as surely as the piston pushing out against a load transfers internal energy from the gas to the external load. In each case given enough time where the ambient temperature is constant, the environment being a giant heat sink/source, the temperatures will equilibriate, and the gas volume will have almost identically the same nRT and therefore PV product as before. That is the static case. The industrial world is based on these simple facts that you reject.
It is funny as an engineer watching you physics guys meassureing small parts of a system after an action within that system has taken place,while over looking everything that took place over the system as a whole after that action. You deal in ideals that dont exist,and messurements are made from these ideals. Then you have variables to make things work out when things dont quite add up. As an engineer,we deal in absolutes,and systems as a whole.
You suffer from a fundamental misconception and have defied all patient efforts to disabuse you of that false notion.
Just found this
Quote
A real process exchanges work with the outside world, specifically, the piston is supposed to do something useful when it moves, the energy to do that cools the gas, just like a can of dusting spray cools when gas is released, and the n*R*T has now gone down inside Tinman's system, meaning that it has given up the energy that performed the external work:
This is a clasic example of what i was saying in the above statement-failing to look at the complete process.-->The total heat energy is still in the system as a whole-->no heat energy has left the system-Elvis is dead,and cannot walk. This is where your 200 year old physics,and your own blindness Mark,MH and the new guru libre are going to come unstuck. This is the very same attitude that holds all man kind back from being free of fossil fuels. 365 days in a year-oh but every 4 years we have to throw another day in there to make things all work out--screwed up with out time definitions there some where-->but hey,what the hell,it's all good.
Now,which of you knows about/or the mathematics to equate total force within a sealed vessel?-->you know,force per unit area x area total. And which of you is able to calculate total heat energy within a given volume? Lets put engineering up against your physics,and we'll see where that lead's us.
What would you like to equate the force to? Force is not energy. The total force is the product of the surface area of the vessel and the pressure. Since for a given internal energy and vessel volume we will have a corresponding pressure that is independent of the surface area of the vessel, for a given internal energy we can design vessels that minimize or maximize force on a given internal surface by our choice of the vessel shape.
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Well it's time you went back to engineering school,as things have changed. We now look at systems as a whole,and all events that will take place around that system in question. Foe example,when meassureing pressure in a sealed vessel,it is no longer acceptable to use psi. You now have to account for total force over a given(total) area.
I don't think I have ever used psi to measure pressure.
Back when I learned engineering the measurement of pressure was "force per unit area". The basic unit in the SI system is the 'Pascal' which is equal to a force of 1 newton per square metre. I realise America is somewhat late to the party but your comment shows extraordinary ignorance for someone who claims to be an engineer.
Now... back to the explanation of why heat and work are convertible and why heat increases when you compress gas, on which you still hold mistaken conclusions.
Thermal energy is the sum of all the kinetic energies of all the particles in the system. When that energy flows between two bodies we call it heat transfer. It is distinct from temperature which is a measure of the potential for heat to flow from one body to another.
To understand why doing work increases the amount of thermal energy in a system and hence the ability to transfer heat, imagine what would happen if we compressed a gas in a cylinder at very high speed, close the actual velocity of the particles in it. In air that would be around 900 km/hr. or approx. the speed of sound.
The piston would then be striking the particles much faster than they normally impact it. That collision imparts a force on the particles and causes their velocity to increase. How could it not..? unless you want to argue the toss on newton's laws of motion.
Since the particles have been accelerated the internal thermal energy increases as it is simply the sum of the kinetic energies of all the particles. Internal energy is increased. The ability is transfer heat is increased.
Of course, the same process occurs at any speed of compression, but it useful to imagine a piston advancing into a gas at high speed to get an intuitive feel for why the velocity and hence 'heat' of those particles increases.
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Mark E says: ' just like a can of
dusting spray cools when gas is released, and the n*R*T has now
gone down inside Tinman's system, meaning that it has given up
the energy that performed the external work'
I say: correct.energy that was supplied by the environment.not by tinman.tinman is merely swtiching phases reversibly >liquid>gas>liquid>gas>liquid>gas.tinman is a traffic cop. Tinman is a transistor switch while the environment does the work
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Mark E says: ' just like a can of
dusting spray cools when gas is released, and the n*R*T has now
gone down inside Tinman's system, meaning that it has given up
the energy that performed the external work'
I say: correct.energy that was supplied by the environment.not by tinman.tinman is merely swtiching phases reversibly >liquid>gas>liquid>gas>liquid>gas.tinman is a traffic cop. Tinman is a transistor switch while the environment does the work
I'm sure there are much better refrigerants than 'HH0', which doesn't actually exist in that form anyway. The gas is H2 and O2 and it combines back to H20. Since that process is at most 70% efficient for a round trip then you'd be far better off using something else as the working fluid.
Like ammonia etc.
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Libre says: 'I'm sure there are much better refrigerants than 'HH0', which
doesn't actually exist in that form anyway. The gas is H2 and O2
and it combines back to H20. Since that process is at most 70%
efficient for a round trip then you'd be far better off using
something else as the working fluid.
Like ammonia etc'
I say: or a singular gas eg h2 alone.at low currents.then we can push it up to 95% reversability and fall within the window of gainland
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I don't think I have ever used psi to measure pressure.
Back when I learned engineering the measurement of pressure was "force per unit area". The basic unit in the SI system is the 'Pascal' which is equal to a force of 1 newton per square metre. I realise America is somewhat late to the party but your comment shows extraordinary ignorance for someone who claims to be an engineer.
Now... back to the explanation of why heat and work are convertible and why heat increases when you compress gas, on which you still hold mistaken conclusions.
Thermal energy is the sum of all the kinetic energies of all the particles in the system. When that energy flows between two bodies we call it heat transfer. It is distinct from temperature which is a measure of the potential for heat to flow from one body to another.
To understand why doing work increases the amount of thermal energy in a system and hence the ability to transfer heat, imagine what would happen if we compressed a gas in a cylinder at very high speed, close the actual velocity of the particles in it. In air that would be around 900 km/hr. or approx. the speed of sound.
The piston would then be striking the particles much faster than they normally impact it. That collision imparts a force on the particles and causes their velocity to increase. How could it not..? unless you want to argue the toss on newton's laws of motion.
Since the particles have been accelerated the internal thermal energy increases as it is simply the sum of the kinetic energies of all the particles. Internal energy is increased. The ability is transfer heat is increased.
Of course, the same process occurs at any speed of compression, but it useful to imagine a piston advancing into a gas at high speed to get an intuitive feel for why the velocity and hence 'heat' of those particles increases.
I can't fault Tinman for using PSI as his preferred unit of pressure. FWIW he is in Australia.
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Libre says: 'I'm sure there are much better refrigerants than 'HH0', which
doesn't actually exist in that form anyway. The gas is H2 and O2
and it combines back to H20. Since that process is at most 70%
efficient for a round trip then you'd be far better off using
something else as the working fluid.
Like ammonia etc'
I say: or a singular gas eg h2 alone.at low currents.then we can push it up to 95% reversability and fall within the window of gainland
95% is not a gain, and it never will be. Why complicate the phase change mechanism when simpler candidates for it already exist.
The overall thermodynamic efficiency will still be constrained by Carnot as well.
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Libre :'95% is not a gain, and it never will be.'
Me: correct.which is where the environment's contributi takes over.over that line.
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Libre :'95% is not a gain, and it never will be.'
Me: correct.which is where the environment's contributi takes over.over that line.
Like a heat pump you mean...? we've had them for a while. They are not over-unity devices.
If you think they possibly can be go back to the explanation of why the heat in a gas increases when it is compressed, and realise you are mistaken.
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Libre:'Like a heat pump you mean...?'
Me: no ordinary heat-pump no.this one can be special
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Libre:'Like a heat pump you mean...?'
Me: no ordinary heat-pump no.this one can be special
No it can't, except perhaps in fantasy land where the heat in a gas remains constant when it is compressed.
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Libre:'fantasy land where the heat in a gas
remains constant when it is compressed.'
Me: welcome to fantasyland where T opposes P (E=RT/nf ln P) during phase-change
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Libre:'fantasy land where the heat in a gas
remains constant when it is compressed.'
Me: welcome to fantasyland where T opposes P (E=RT/nf ln P) during phase-change
That is of no consequence. He believes the heat in a gas does not change when work is done on it. It does change and has been described many times why that is so. If it worked as he proposes then yes it would be over-unity. You could create a 'cold sink' and cause heat from the environment to flow into it without having to exhaust it back to the environment against a temperature gradient.
Unless of course you believe the phase change event can also exhibit net energy gain around a full cycle , then it's game over for this device.
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I can't fault Tinman for using PSI as his preferred unit of pressure. FWIW he is in Australia.
In post #142 he apparently has a problem with using 'psi' as a unit of pressure measurement.
I was just pointing out I never have. Any engineering ever done by myself was in SI units.
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In post #142 he apparently has a problem with using 'psi' as a unit of pressure measurement.
I was just pointing out I never have. Any engineering ever done by myself was in SI units.
While I greatly prefer to work in SI units, scaling is just nuisance arithmetic. In 142 he seemed to be saying he thinks it is necessary to integrate to find total force. That would make sense if one were trying to find stress on a joint.
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While I greatly prefer to work in SI units, scaling is just nuisance arithmetic. In 142 he seemed to be saying he thinks it is necessary to integrate to find total force. That would make sense if one were trying to find stress on a joint.
I'm not entirely sure what he meant. In an ideal gas at a single point the forces are exerted equally in all directions so it is sufficient to use force divided by area. I've never come across having to use the integral in most engineering analysis.
Perhaps atmospheric processes where the pressure varies with altitude would be an example where you would.
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I'm not entirely sure what he meant. In an ideal gas at a single point the forces are exerted equally in all directions so it is sufficient to use force divided by area. I've never come across having to use the integral in most engineering analysis.
Perhaps atmospheric processes where the pressure varies with altitude would be an example where you would.
He might be thinking of the case of a vessel filled with dense fluid and not a low density gas.
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I'm sure there are much better refrigerants than 'HH0', which doesn't actually exist in that form anyway. The gas is H2 and O2 and it combines back to H20. Since that process is at most 70% efficient for a round trip then you'd be far better off using something else as the working fluid.
Like ammonia etc.
Libre
You really need to go back to the start a d read the thread, and the accounted for energy during the electrolysis process. The system is designed to produce heat, and it dose that at 100% efficiancy-all systems are 100% efficient when all output energies are accounted for.
We do work with units of force, but to simplify we use the psi units-either is correct.
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Libre
You really need to go back to the start a d read the thread, and the accounted for energy during the electrolysis process. The system is designed to produce heat, and it dose that at 100% efficiancy-all systems are 100% efficient when all output energies are accounted for.
We do work with units of force, but to simplify we use the psi units-either is correct.
I just don't see the point in using electrolysis when another substance undergoing a phase change would be more efficient. It is just unnecessary complication. If it designed to produce heat as you say then the co-efficient of performance would be higher using something else.
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Tinman:
I have been doing this stuff in my head mostly, trying to rely on my wits. I have looked at formulas from time to time but it's mostly from thinking through the problem.
Going back to the transparent cylinder with the piston, I said that if you let the heat escape from the compressed gas first, and then let the piston return by itself, it would not return all the way back in one stroke. It would slowly creep back last 1/4 of the way. We are assuming the plunger is frictionless, but otherwise it's a real bench setup.
Do you agree with that? If yes, when the piston stops after moving 3/4 of the way back to the start position, what is the temperature of the gas?
MileHigh
MH
I ask once again that you read the thread from the star and study the system setup. The system is in an insulated room, and when opperating temperatures are reached there is no temperature difference between the gas inside the cylinder and the external enviromental temperature-the heat simply cannot dissipate-Elvis simply can not leave the building-the door is locked-the system is an enclosed system, and the energy remains within that system. It comes down to this-do you want to look at what is really taking place within my system, or do you want to play with a cylinder that is open to the enviroment where its internal energy can be transfered to. Remember MH, energy can be transfered or transformed-not just up and disappear.
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I ask once again that you read the thread from the star and study the system setup. The system is in an insulated room, and when opperating temperatures are reached there is no temperature difference between the gas inside the cylinder and the external enviromental temperature-the heat simply cannot dissipate-Elvis simply can not leave the building-the door is locked-the system is an enclosed system, and the energy remains within that system. It comes down to this-do you want to look at what is really taking place within my system, or do you want to play with a cylinder that is open to the enviroment where its internal energy can be transfered to. Remember MH, energy can be transfered or transformed-not just up and disappear.
The problem is that you still appear to believe that using work to compress something does not raise the amount of thermal energy or 'heat' it contains. It does and until you understand why you'll continue to believe in this fantasy device
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Libre :'Unless of course you believe the phase change event can also
exhibit net energy gain around a full cycle , then it's game over
for this device.'
Me: at low current density/ electrode areas your going to get this if T lowers E greater than P increases E or even if it just cancells it and E remains unchanged during this cycle
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I just don't see the point in using electrolysis when another substance undergoing a phase change would be more efficient. It is just unnecessary complication. If it designed to produce heat as you say then the co-efficient of performance would be higher using something else.
The cop with this system is 1. This i have been through with Mark-energy is conserved-->are you going to argue with that?.
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While I greatly prefer to work in SI units, scaling is just nuisance arithmetic. In 142 he seemed to be saying he thinks it is necessary to integrate to find total force. That would make sense if one were trying to find stress on a joint.
Do you know of any pressure vessels that have no joint?.
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Thermal energy is the sum of all the kinetic energies of all the particles in the system. When that energy flows between two bodies we call it heat transfer. It is distinct from temperature which is a measure of the potential for heat to flow from one body to another.
Since the particles have been accelerated the internal thermal energy increases as it is simply the sum of the kinetic energies of all the particles. Internal energy is increased. The ability is transfer heat is increased.
I don't think I have ever used psi to measure pressure
.
Well here in Australia it is the most commonly used messure of force. There are times when the term force is used,but force is pressure-so either is ok.
but your comment shows extraordinary ignorance for someone who claims to be an engineer.
why/-because it dosnt agree with what you believe or the units you may use where ever you are. America still uses inches and feet,while in australia we use mm and meters-dose that make us all ignorant,or are we just a little ahead of the American's?
Of course, the same process occurs at any speed of compression, but it useful to imagine a piston advancing into a gas at high speed to get an intuitive feel for why the velocity and hence 'heat' of those particles increases.
You think that faster moving particals create more heat,where as i think more heat causes the particles to move faster. What happens to the particles in the air when the air is warmed by the sun-->the particles move faster. Now there was no compressing of the particles there,and yet they sped up.
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What would you like to equate the force to?
Yes, you are very badly lost on very fundamental issues.
We shall see soon enough.
Wow Mark,you have just invented an OU device. We now have more stored force than we started with. How did you do that when it was clearly stated that it was a frictionless device.
No, this is again where you are just lost. Please avail yourself to any of the many thousands of reputable online references because you stopped paying attention to any of the correct explanations offered by others and myself here a long time ago.
You have to understand that i dont see all post when im working 17 hours a day,they are easly missed when trying to keep up during work hours. I am interested as to how you managed to achieve a higher pressure after all the energy was returned on the return stroke of the piston.
You have made a series of incorrect statements that are refuted left and right by that "200 year old" physics that you disdain. Experiment after experiment refutes your assertions.
You mean you think i have made a series of incorrect statements based on your belief that the ideal gas laws can be applied to my system.
My definition is one who engages in personal attack rather than debate the merits of their arguments.Yes it is!" and "No it isn't!" are declarations, not arguments.
Like telling some one who has a different belief than yourself that there full of shit?-Maybe that was just your way of telling me no.
For the umpteenth time, if the piston does work on the outside world, then it transfers energy to the outside world.
Respectfully !no!-no energy is lost to the outside world,as the unit as a whole is all enclosed in an insulated room.
It exerts that force in the direction of motion. It performs work. That work comes from the internal energy of the gas.
No,that work comes from the force/pressure that already exist and is already accounted for within the system.
Going the opposite way: External force must be applied to the piston in order to force the gas into a smaller volume.
Not part of the process in my system,as the gas is already compressed.
That work increases the internal energy of the gas. It increases nRT and consequently the PV product just as surely as the piston pushing out against a load transfers internal energy from the gas to the external load
.
No,the gas and all the energy it contains remains within the internal system. You just dont see the opperation of the system as a whole because you believe that your ideal gas law applies to all the action within my system.
The industrial world is based on these simple facts that you reject.You suffer from a fundamental misconception and have defied all patient efforts to disabuse you of that false notion.
Once again,you assume that the ideal gas laws can be applied to my system-they cannot while the system is in opperation.
The total force is the product of the surface area of the vessel and the pressure.
Indeed-on this we agree.Please remember you agree with this,as it will come into play soon enough.
Force is not energy. Since for a given internal energy and vessel volume we will have a corresponding pressure that is independent of the surface area of the vessel, for a given internal energy we can design vessels that minimize or maximize force on a given internal surface by our choice of the vessel shape.
No ,force is not energy,but gives rise to energy.
The vessel of choice in this system will be a cylinder type vessel.
Mark.
You think i show you disrespec,but that is what i have seen from you toward myself. The button was pushed when you said i was full of shit just because i have different beliefs to you. From then on you were given as good as you gave,and there are many examples of that in this thread. You telling me i need to go do some study,and me telling you you need to stop living in the past-you telling me i dont understand or know what im doing,and me telling you that you dont understand my system,and that your ideal gas laws cannot be applied to my system while it is in operation. It can go on and on,and at the moment im guessing the one thing we gan agree on is that neither of us has respec for the other. I am not sure what you feel that your position here on this forum is,but im thinking that you see your self as some sort of teacher,and we are your student's. Well im sorry,but im not here to live by the ways of old-->im here to find the holes in the physics that you love so much. Im here to find the errors in those said laws,and to find where they dont apply-->and i believe my system is one such case.
So lets start working on this together,insted of seeing who can insult the other the most. First up,some question's.
1-As i understand,there has to be one constant for us to be able to use the ideal gas laws on my system. Either P,V,T or n--Is this correct?
2- If the pressure rises within the storage tank/vessel,then the heat will rise as well?
3-If the volume increases,but n is constant,then the pressure and heat will drop?
And 4-if i give you the parameters,are you able to convert the energy stored within the storage vessel into joules of energy using the ideal gas laws?.
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The problem is that you still appear to believe that using work to compress something does not raise the amount of thermal energy or 'heat' it contains. It does and until you understand why you'll continue to believe in this fantasy device
It is you that misunderstands libre. It is heat that causes the gas molecules to move faster,and this means that they will hit the walls of the vessel faster. This is why you then get an increase in pressure.
Quote: In gases, heat allows the gases particles to move faster. They will hit the walls of the container with more force. If the walls are allowed to move, then the walls will actually move to a bigger volume because of the much stronger striking of the gas particles against the walls. Again the container becomes bigger.
Quote science learn.org--An effect of heat – expansion
When gases, liquids and solids are heated, they expand. As they cool, they contract or get smaller. The expansion of the gases and liquids is because the particles are moving around very fast when they are heated and are able to move further apart so they take up more room. If the gas or liquid is heated in a closed container, the particles collide with the sides of the container, and this causes pressure. The greater the number of collisions, the greater the pressure.
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@Tinman
MOST readers seem to be understanding your concept without too much
dificulty.
It may work out It may not. Personally, I think its worth something.
Remember some people here are just want to pull on people's chain.
Basics
We Know temperature <> total heat energy.
When a force is applied to cause compression of a gas
that Kinetic energy so added to the gas is the cause of the temperature
rise. But that Kinetic energy is (not usually significantly) transformed into heat
energy. But remains in Kinetic energy form and is stored as potential energy / pressure.
Total energy is increased, but total thermal energy in not significantly increased (usually).
Sonic velocities ???
Diagram included (Tinman Open sys1. pdf)
cheers floor
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Libre says to tinman: 'The problem is that you still appear to believe that using work to
compress something does not raise the amount of thermal energy
or 'heat' it contains. It does and until you understand why you'll
continue to believe in this fantasy device'
I say to libre: that heat does come into that gas but not from tinman.it comes from the environment.free of charge
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Libre :'I just don't see the point in using electrolysis when another
substance undergoing a phase change would be more efficient.'
Me: because phase change during electrolysis does not exclusively tax the operator
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The cop with this system is 1. This i have been through with Mark-energy is conserved-->are you going to argue with that?.
Energy is conserved but you are not accounting for it correctly.
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Mark E says to tinman: 'Energy is conserved but you are not accounting for it correctly.'
I say to mark E: he's on track
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.
Well here in Australia it is the most commonly used messure of force. There are times when the term force is used,but force is pressure-so either is ok.
The difference between pressure and force is so basic that I hope you mean you do something other than substitute pressure directly for force.
...
You think that faster moving particals create more heat,where as i think more heat causes the particles to move faster. What happens to the particles in the air when the air is warmed by the sun-->the particles move faster. Now there was no compressing of the particles there,and yet they sped up.
LE's point is that faster moving particles have more internal energy. That is simple kinetics. The industrial world operates on the fact that gasses can exchange internal energy with the outside world by either thermal or mechanical means.
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Quote from: MarkE on February 04, 2015, 09:01:52 PM
What would you like to equate the force to?
Quote
Yes, you are very badly lost on very fundamental issues.
We shall see soon enough.
Quote
Wow Mark,you have just invented an OU device. We now have more stored force than we started with. How did you do that when it was clearly stated that it was a frictionless device.
The good news is that you intend to experiment. The bad news is that at least for the time being you remain at odds with the very basics of thermodynamics for which there are thousands of good tutorials available to you for free on the www.
No, this is again where you are just lost. Please avail yourself to any of the many thousands of reputable online references because you stopped paying attention to any of the correct explanations offered by others and myself here a long time ago.
You have to understand that i dont see all post when im working 17 hours a day,they are easly missed when trying to keep up during work hours. I am interested as to how you managed to achieve a higher pressure after all the energy was returned on the return stroke of the piston.
I appreciate that you work very hard.
Quote
You have made a series of incorrect statements that are refuted left and right by that "200 year old" physics that you disdain. Experiment after experiment refutes your assertions.
You mean you think i have made a series of incorrect statements based on your belief that the ideal gas laws can be applied to my system.
No, with all due respect: We are talking about very basic, extraordinarily well proven principles. While anyone can be mistaken, and any principle has a finite possibility of being wrong under some circumstance: the chances here are so close to nil that I make no distinction for the extremely good cause that there is no evidence of any kind that disputes these fundamentals which have been proven billions of times by every hot gas engine and heat pump built over the past 200 years.
Quote
My definition is one who engages in personal attack rather than debate the merits of their arguments.Yes it is!" and "No it isn't!" are declarations, not arguments.
Like telling some one who has a different belief than yourself that there full of shit?-Maybe that was just your way of telling me no.
The distinction that I draw is decrying ideas that cannot stand rather than the person. If you feel personally offended by anything I have said then I apologize for that.
Quote
For the umpteenth time, if the piston does work on the outside world, then it transfers energy to the outside world.
Respectfully !no!-no energy is lost to the outside world,as the unit as a whole is all enclosed in an insulated room.
Here we are back at the crux of the situation. Is it fair to say that you believe that because at a given internal energy a particular quantity of gas has a P*V product that P or V can be changed to new values without usng or adding energy so long as the P*V product remains fixed? If so that is called an adiabatic condition. It is an extremely useful, but hypothetical condition. If you understand that work is the integral of F*ds, or less rigorously: force times distance, then you should understand that if we have gas in a cylinder, and if we change the volume of the cylinder, then we apply force along the axis of the cylinder and therefore either put work into the cylinder by compressing it, or take work out by allowing the cylinder to expand. LE has used the bicycle pump example and I have used the aerosol can example. I don't think that there is any debate that where P1V1 = P2V2 the internal energy is the same. The debate is whether it is possible to change state from P1V1 without exchanging external energy.
Quote
It exerts that force in the direction of motion. It performs work. That work comes from the internal energy of the gas.
No,that work comes from the force/pressure that already exist and is already accounted for within the system.
See above.
Quote
Going the opposite way: External force must be applied to the piston in order to force the gas into a smaller volume.
Not part of the process in my system,as the gas is already compressed.
Again, see above.
Quote
That work increases the internal energy of the gas. It increases nRT and consequently the PV product just as surely as the piston pushing out against a load transfers internal energy from the gas to the external load
.
No,the gas and all the energy it contains remains within the internal system. You just dont see the opperation of the system as a whole because you believe that your ideal gas law applies to all the action within my system.
See above.
Quote
The industrial world is based on these simple facts that you reject.You suffer from a fundamental misconception and have defied all patient efforts to disabuse you of that false notion.
Once again,you assume that the ideal gas laws can be applied to my system-they cannot while the system is in opperation.
Your first problem is that if you want to claim an exception to general principles then you either need to make an argument for why that exception should exist, or demonstrate that it appears to exist.
Quote
The total force is the product of the surface area of the vessel and the pressure.
Indeed-on this we agree.Please remember you agree with this,as it will come into play soon enough.
That's not a problem for me.
Quote
Force is not energy. Since for a given internal energy and vessel volume we will have a corresponding pressure that is independent of the surface area of the vessel, for a given internal energy we can design vessels that minimize or maximize force on a given internal surface by our choice of the vessel shape.
No ,force is not energy,but gives rise to energy.
The vessel of choice in this system will be a cylinder type vessel.
Energy is the integral of F*ds. We must have force and some portion of it must be applied through the direction of motion.
Mark.
You think i show you disrespec,but that is what i have seen from you toward myself. The button was pushed when you said i was full of shit just because i have different beliefs to you. From then on you were given as good as you gave,and there are many examples of that in this thread. You telling me i need to go do some study,and me telling you you need to stop living in the past-you telling me i dont understand or know what im doing,and me telling you that you dont understand my system,and that your ideal gas laws cannot be applied to my system while it is in operation. It can go on and on,and at the moment im guessing the one thing we gan agree on is that neither of us has respec for the other. I am not sure what you feel that your position here on this forum is,but im thinking that you see your self as some sort of teacher,and we are your student's. Well im sorry,but im not here to live by the ways of old-->im here to find the holes in the physics that you love so much. Im here to find the errors in those said laws,and to find where they dont apply-->and i believe my system is one such case.
I hope you appreciate that we did not arrive at our current understanding of the physical world by proclamation. Our understanding is based on countless experiments devised and executed by very bright people. While there is always more to learn, and there is always the finite possibility that we will find an exception to basic principles, the odds against any kind of experiment that repeats conditions that have been tested millions of times finding something new are really, really, small.
So lets start working on this together,insted of seeing who can insult the other the most. First up,some question's.
1-As i understand,there has to be one constant for us to be able to use the ideal gas laws on my system. Either P,V,T or n--Is this correct?
No, not really. In order for the ideal gas law to be a good approximation the effects of things like Van Der Waal forces have to be small. nRT is going to give the internal energy of the gas. Under any static conditions and volumes like you are talking about PV will be so close to nRT that we do not distinguish the two.2- If the pressure rises within the storage tank/vessel,then the heat will rise as well?
A higher P*V product means more internal energy, IE higher nRT, than a lower P*V product. R is a constant. So if n is constant, then P*V changes with T and vice-versa.3-If the volume increases,but n is constant,then the pressure and heat will drop?
The internal energy is nRT. The disagreement that you have with basic thermodynamics is the idea that you can change P or V without exhanging work that would change nRT and therefore the product of P*V. You are free to try and find a way to enlarge V and still keep Elvis in the building.And 4-if i give you the parameters,are you able to convert the energy stored within the storage vessel into joules of energy using the ideal gas laws?.
It's a simple calculator plug to take n and T and multiply by R. If n is unknown, but P and V are known, it is also a simple calculator plug to get the energy.
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You think that faster moving particals create more heat,where as i think more heat causes the particles to move faster. What happens to the particles in the air when the air is warmed by the sun-->the particles move faster. Now there was no compressing of the particles there,and yet they sped up.
This is incorrect.
Heat is the sum of the kinetic energies of all the particles in the system. A system being heated doesn't "care" how the particles in it were accelerated, only that they were. If they are going faster they have more heat.
The increased movement might have been caused by absorbing radiation or physical collisions with another particle. In the case of IR absorption some is re-emitted, but a portion is absorbed and causes the particle to move faster.
In the case of a piston advancing into a cylinder the piston is hitting particles faster than when it is at rest, and that momentum is transferred to the particle. After the piston stops the gas now has more internal energy or heat. It is as simple as that.
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The cop with this system is 1. This i have been through with Mark-energy is conserved-->are you going to argue with that?.
If the COP is 1 then an electrical resistance heater would be a much simpler device to achieve the same result.
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We shall see soon enough.
Quote
Wow Mark,you have just invented an OU device. We now have more stored force than we started with. How did you do that when it was clearly stated that it was a frictionless device.
The good news is that you intend to experiment. The bad news is that at least for the time being you remain at odds with the very basics of thermodynamics for which there are thousands of good tutorials available to you for free on the www.
You have to understand that i dont see all post when im working 17 hours a day,they are easly missed when trying to keep up during work hours. I am interested as to how you managed to achieve a higher pressure after all the energy was returned on the return stroke of the piston.
I appreciate that you work very hard.
You mean you think i have made a series of incorrect statements based on your belief that the ideal gas laws can be applied to my system.
No, with all due respect: We are talking about very basic, extraordinarily well proven principles. While anyone can be mistaken, and any principle has a finite possibility of being wrong under some circumstance: the chances here are so close to nil that I make no distinction for the extremely good cause that there is no evidence of any kind that disputes these fundamentals which have been proven billions of times by every hot gas engine and heat pump built over the past 200 years.
Like telling some one who has a different belief than yourself that there full of shit?-Maybe that was just your way of telling me no.
The distinction that I draw is decrying ideas that cannot stand rather than the person. If you feel personally offended by anything I have said then I apologize for that.
Respectfully !no!-no energy is lost to the outside world,as the unit as a whole is all enclosed in an insulated room.
Here we are back at the crux of the situation. Is it fair to say that you believe that because at a given internal energy a particular quantity of gas has a P*V product that P or V can be changed to new values without usng or adding energy so long as the P*V product remains fixed? If so that is called an adiabatic condition. It is an extremely useful, but hypothetical condition. If you understand that work is the integral of F*ds, or less rigorously: force times distance, then you should understand that if we have gas in a cylinder, and if we change the volume of the cylinder, then we apply force along the axis of the cylinder and therefore either put work into the cylinder by compressing it, or take work out by allowing the cylinder to expand. LE has used the bicycle pump example and I have used the aerosol can example. I don't think that there is any debate that where P1V1 = P2V2 the internal energy is the same. The debate is whether it is possible to change state from P1V1 without exchanging external energy.
No,that work comes from the force/pressure that already exist and is already accounted for within the system.
See above.
Not part of the process in my system,as the gas is already compressed.
Again, see above.
No,the gas and all the energy it contains remains within the internal system. You just dont see the opperation of the system as a whole because you believe that your ideal gas law applies to all the action within my system.
See above.
Once again,you assume that the ideal gas laws can be applied to my system-they cannot while the system is in opperation.
Your first problem is that if you want to claim an exception to general principles then you either need to make an argument for why that exception should exist, or demonstrate that it appears to exist.Indeed-on this we agree.Please remember you agree with this,as it will come into play soon enough.
That's not a problem for me.
No ,force is not energy,but gives rise to energy.
The vessel of choice in this system will be a cylinder type vessel.
Energy is the integral of F*ds. We must have force and some portion of it must be applied through the direction of motion.
Mark.I hope you appreciate that we did not arrive at our current understanding of the physical world by proclamation. Our understanding is based on countless experiments devised and executed by very bright people. While there is always more to learn, and there is always the finite possibility that we will find an exception to basic principles, the odds against any kind of experiment that repeats conditions that have been tested millions of times finding something new are really, really, small.No, not really. In order for the ideal gas law to be a good approximation the effects of things like Van Der Waal forces have to be small. nRT is going to give the internal energy of the gas. Under any static conditions and volumes like you are talking about PV will be so close to nRT that we do not distinguish the two.A higher P*V product means more internal energy, IE higher nRT, than a lower P*V product. R is a constant. So if n is constant, then P*V changes with T and vice-versa.The internal energy is nRT. The disagreement that you have with basic thermodynamics is the idea that you can change P or V without exhanging work that would change nRT and therefore the product of P*V. You are free to try and find a way to enlarge V and still keep Elvis in the building.It's a simple calculator plug to take n and T and multiply by R. If n is unknown, but P and V are known, it is also a simple calculator plug to get the energy.
Its not going to be as simple as you assume Mark, in fact this will be a test for you indeed-I hope your maths is up to scratch.
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If the COP is 1 then an electrical resistance heater would be a much simpler device to achieve the same result.
Good one LE.
Now how would a resistance heater produce pressure of around 400psi-which is yet to come.
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Good one LE.
Now how would a resistance heater produce pressure of around 400psi-which is yet to come.
All it has to do is boil a liquid in a suitable pressure vessel.
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Here's a graph of tinmans engine losses vs environmental donations.the environment will donate a set number of kcal/mole heat for phase-changes whilst tinman must make sure his own losses don't exceed that set number in the cycle.
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All it has to do is boil a liquid in a suitable pressure vessel.
An electrolisis cell is a resistance heater-->and we get gas as well ;)
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An electrolisis cell is a resistance heater-->and we get gas as well ;)
Electrolysis and recombination of water is a lossy process. Expanding a gas to produce work is a lossy process. The environment cannot contribute net energy as this would require a cold sink at lower temperature than the environment. The combination of 2 or more lossy processes multiplies the possible losses. I'm looking hard but I just don't see any useful process here.
Now if by chance work did not raise the amount of heat in a gas I'd be forced to reconsider my position.. All you need to prove is it does not and I'd happily endorse this...
Waiting for proof now :)
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Electrolysis and recombination of water is a lossy process. Expanding a gas to produce work is a lossy process. The environment cannot contribute net energy as this would require a cold sink at lower temperature than the environment. The combination of 2 or more lossy processes multiplies the possible losses. I'm looking hard but I just don't see any useful process here.
Now if by chance work did not raise the amount of heat in a gas I'd be forced to reconsider my position.. All you need to prove is it does not and I'd happily endorse this...
Waiting for proof now :)
There is no loss when you account for all the energy output of any system-energy must be conserved-->this is one law i do agree with.
The energy stored within the H2O1 gas,plus the heat energy is 100% of the input energy-->COP1.
With the addition of pressure,there are now three energy sources-disipated heat,energy stored within the gas itself,and the pressure-->which in turn is heat energy. The more we raise the pressure,the more heat the cell will produce. This is what i am trying to explain here-the pressure has already been accounted for in the form of heat energy,and we can use this pressure without added cost to the devices input energy.
Anyway,we will go through it step by step,and messure each of the two gas storage tank's energy as the process takes place-providing Mark can nut out the math involved,which is going to be quite complex.
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Libre says:'The environment cannot contribute net energy as this would require a cold sink at lower temperature than the environment'
I say: yes a phase-change sink like h2o> h2 + o2.colder than the environment
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libre:'electrolysis and recombination of water is a lossy process. Expanding a gas to produce work is a lossy process. The environment cannot contribute net energy'
Me: depends on your setup
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An electrolisis cell is a resistance heater-->and we get gas as well ;)
Sure, and the energy out all comes from the battery just as the energy out from the resistance heater does. So in each case energy has been delivered to gas inthe pressure vessel. In the case of the electrolysis, in order to get the pressure up in the cell you need to liberate the energy used to break the chemical bonds. Assuming you have really good insulation, the results would be essentially the same.
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Mark:
So, 1 for 1 then? Or possibly a little less than one out so maybe 98%?
Bill
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Mark:
So, 1 for 1 then? Or possibly a little less than one out so maybe 98%?
Bill
Ultimately we will have to see what kind of experiment Tinman has in mind.
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Ultimately we will have to see what kind of experiment Tinman has in mind.
Im leaving on vacation tomorrow,as it's been 20 years in the making.
But i leave you with some thoughts to be given when we go to do our test Mark.
the primary storage tank wont be that hard,but the ram is where math skills will be needed-and good ones at that.
If you look at the diagram on page one,you will see what and how the ram is set up. You will note that it has a compression spring in it. This changes everything that happens when the cylinder is opperating.
The pressure will continue to rise over the period of the stroke due to the springs required compression force continually rising to compress it. The volume will also be continually rising,and also the gas mass will be continually rising.
As the pressure is continually rising,then so will the teperature continually rise. as the temperature is continually rising,the gas will be expanding along with the rise in temperature. This will reduce the required flow from the main tank. The temperature will end up being higher than that of the gas being delivered to the ram via the tank due to the increasing pressure rise within the cylinder-even though the end pressure in the cylinder will be lower than that of the main tank pressure.
This is your adiabatic process at work Mark.
So you see,the spring is actually going to raise the energy stored within the pneumatic cylinder-not decrease it,and any other load placed on the cylinder increases the rate of the pressure and temperature increase-->the end temperature will be higher with the spring and load on the cylinder than if there were no load at all.
So in summary.
The volume is continually increasing.
The pressure is continually increasing.
The temperature is continually increasing.
The gas mass is continually increasing.
The springs force is continually increasing.
= an increase in energy stored within the cylinder to that supplied by the main tank.
The main tank/storage tank.
The volume remains constant.
The pressure will drop,but will not be linear to that of the cylinder/ram.
The temperature will drop,but will not be linear to that of the cylinder/ram.
The gas mass will be reduced,and will be linear to that of the cylinders mass increase-->we cannot create mass.
These are the things that must be taken into account when on paper testing is carried out.
We are also going to limit the pressure to the cylinder/ram to 100 psi. We will be using psi gage pressure for simplicity.
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Im leaving on vacation tomorrow,as it's been 20 years in the making.
But i leave you with some thoughts to be given when we go to do our test Mark.
the primary storage tank wont be that hard,but the ram is where math skills will be needed-and good ones at that.
If you look at the diagram on page one,you will see what and how the ram is set up. You will note that it has a compression spring in it. This changes everything that happens when the cylinder is opperating.
The pressure will continue to rise over the period of the stroke due to the springs required compression force continually rising to compress it. The volume will also be continually rising,and also the gas mass will be continually rising.
As the pressure is continually rising,then so will the teperature continually rise. as the temperature is continually rising,the gas will be expanding along with the rise in temperature.
When the A valve is first opened the gas expands against the spring. If the dimensions are established that equilibrium condition is not difficult to resolve. If the valve is left open thent he cylinder continues to compress the spring at which point: P, V, and T all increase. This will reduce the required flow from the main tank. The temperature will end up being higher than that of the gas being delivered to the ram via the tank due to the increasing pressure rise within the cylinder-even though the end pressure in the cylinder will be lower than that of the main tank pressure.
This is your adiabatic process at work Mark.
The process is not adiabatic. Once you provide dimensions, then we can do the math.
So you see,the spring is actually going to raise the energy stored within the pneumatic cylinder-not decrease it,and any other load placed on the cylinder increases the rate of the pressure and temperature increase-->the end temperature will be higher with the spring and load on the cylinder than if there were no load at all.
No, the spring stores energy transferred from the gas.
So in summary.
The volume is continually increasing.
Only after the A valve is opened.The pressure is continually increasing.
Only after the equilibriation that occurs after the A valve is opened and the cylinder reaches a stable position. Between the time that the A valve is opened and equilibrium is reached, P, and PV, and T all decrease.The temperature is continually increasing.
No, same thing as the pressure: An initial decrease occurs when A valve is opened until equilibrium is reached. Then the temperature begins rising again.The gas mass is continually increasing.
Yes, we keep converting liquid to gas through the whole electrolysis phase.The springs force is continually increasing.
Once the A valve is opened the spring gets compressed more and more until the electrolysis ends.= an increase in energy stored within the cylinder to that supplied by the main tank.
That's right, we transfer energy from the main tank to the cylinder from the time the A valve opens until we stop the electrolysis.
The main tank/storage tank.
The volume remains constant.
Absolutely not. It increases until we open valve A, then it drops until equilibrium is reached and then it starts increasing again.The pressure will drop,but will not be linear to that of the cylinder/ram.
The pressure will drop until the pressure supplied by the newly compressed (more compressed) spring against the piston reaches equilibrium with the pressure in the vessel + expanded cylinder. Energy transferred to the spring drops nRT and consequently P*V.The temperature will drop,but will not be linear to that of the cylinder/ram.
The temperature conforms to the internal energy of the gas.The gas mass will be reduced,and will be linear to that of the cylinders mass increase-->we cannot create mass.
Unless gas escapes, the total mass of gas only goes up as the electrolysis proceeds. We can't create mass, but we can generate gas from liquid.
These are the things that must be taken into account when on paper testing is carried out.
The paper model better be faithful to real physics or it won't be very useful.We are also going to limit the pressure to the cylinder/ram to 100 psi. We will be using psi gage pressure for simplicity.
How do you intend to limit the pressure?
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Closed systems are essentially non existent its best to call them closed loops within open systems in this respect every atom is electrically connected to one another by means of electric charges as well as gravitational force(s) and interaction between the energy flowing in closed loops within open systems and the energy flowing in open systems only becomes apparent when there is enough potential or disequilibrium available to establish a noticeable conductive pathways between said systems in order to make use of natures re balancing flux or better said to invite the external energy in(on)to the closed loop within the open system it is not overunity nor is it free energy or anything mysterious when all the available energy and conversions are taken into account.
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Turbo
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Closed systems are essentially non existent
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very true , and an important perspective .
Chet
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Turbo
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Closed systems are essentially non existent
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very true , and an important perspective .
Chet
Closed systems 'exist'. They are simply a convenient way of analysing an idealised process mathematically.
Sure, they don't exist in the 'real' world, as we can only approximate them. importantly however the results of the idealised analysis puts absolute limits on the efficiency of real world processes.
Carnot for instance tell us that for any conversion of heat to work that the absolute limit on the efficiency is proportional to the temperature difference between the hot and cold sink and you cannot invent a real world process that breaks that rule.
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libre:'and you cannot invent a real world process that breaks that rule.'
Me: are you sure about this
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libre:'and you cannot invent a real world process that breaks that rule.'
Me: are you sure about this
For systems comprised of gasses that are modelled by the ideal gas law... absolutely.
For systems comprising mainly components where quantum scale effects predominate, possibly not.
However it is futile to build a system where all of the components are larger than quantum sized and expect it to behave any differently than the long established thermodynamic laws.
For sure these laws have only a statistical certainty but the chances of them behaving differently than how we know they apply in the normal physical world is vanishingly small.
However, IF you were to propose a system however where the geometry meant that the ratio of electromagnetic to gravitational forces changed (such as inside a Casmir cavity perhaps) then possibly yes those laws may no longer apply.
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Libre:'However, IF you were to propose a system however where the geometry meant that the ratio of electromagnetic to gravitational forces changed (such as inside a Casmir cavity perhaps) then possibly yes those laws may no longer apply.'
Me: for example a catalyst system.electrolysis and catalysts tend to go well together you agree?
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Unfortunately MarkE or LE just dont see or understand what is happening within my system as a whole. How exactly they think the energy containd within the gas is going to leave the system is yet to be explained.
Neither heat or gas mass leaves the system, and contrary to Marks belief, the spring actually increases the heat energy within the gas in the cylinder-this will be shown soon enough.
There is simply no way for the energy within the gas to drop when used as it is in my system.
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Unfortunately MarkE or LE just dont see or understand what is happening within my system as a whole. How exactly they think the energy containd within the gas is going to leave the system is yet to be explained.
Neither heat or gas mass leaves the system, and contrary to Marks belief, the spring actually increases the heat energy within the gas in the cylinder-this will be shown soon enough.
There is simply no way for the energy within the gas to drop when used as it is in my system.
I have asked you to confirm my understanding of your beliefs. Is it your belief that you can take a volume V1 of some gas at some pressure P1 and without adding or subtracting energy from that gas volume, change the volume? IE end up with a P2*V2 = P1*V1 where P1<>P2? I think that is your belief. If it is, kindly show an example of how you can get between two pressure values without either adding or removing energy. Please show your work.
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Tinman is theoreticly arguing from a 100% reversability pov @mark E.like in this diagram.if twere 100% reversable,would there be a gain in work out after pressure increase.we use two seperate solid ion-exchange membranes.one for electrolysis.one for galvanosis
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As much as id like to Mark, im on vacation. But I must ask-what energy do you think is going to leave the gas when the piston is opperated?
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As much as id like to Mark, im on vacation. But I must ask-what energy do you think is going to leave the gas when the piston is opperated?
Are you clueless? it would seem so. What happens has been described to you many times already.
Work is converted to heat and vice versa.
As the piston advances it hits the particles with greater momentum than when it is at rest. Since the collisions are fairly elastic, most of that momentum is transferred to the gas particles making them go faster. Faster particles means more heat. Fairly simple really. The opposite happens when decompressing the gas.
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Are you clueless? it would seem so. What happens has been described to you many times already.
Work is converted to heat and vice versa.
As the piston advances it hits the particles with greater momentum than when it is at rest. Since the collisions are fairly elastic, most of that momentum is transferred to the gas particles making them go faster. Faster particles means more heat. Fairly simple really. The opposite happens when decompressing the gas.
It is also rather convenient, because it's what has been driving the industrial world for 200 years.
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Are you clueless? it would seem so. What happens has been described to you many times already.
Work is converted to heat and vice versa.
As the piston advances it hits the particles with greater momentum than when it is at rest. Since the collisions are fairly elastic, most of that momentum is transferred to the gas particles making them go faster. Faster particles means more heat. Fairly simple really. The opposite happens when decompressing the gas.
It is you that is clueless.
Im refering to the gas pushing the piston down, not the piston compressing the gas. This brings the point that you simply dont understand equal and opposites. If the gas is pushing on the piston, then the piston is also pushing on the gas just as hard. So as the pressure bjilds, then so dose the heat. So, the heavier the load, the faster the pressure builds.
Do you not understand equal and opposite reactions LE?
So I ask once again-what energy do you think is going to leave the gas when that gas is used to force the piston down?
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Please consult the diagram above gentlmen.its the same fucking thing
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It is you that is clueless.
Im refering to the gas pushing the piston down, not the piston compressing the gas. This brings the point that you simply dont understand equal and opposites. If the gas is pushing on the piston, then the piston is also pushing on the gas just as hard. So as the pressure bjilds, then so dose the heat. So, the heavier the load, the faster the pressure builds.
Do you not understand equal and opposite reactions LE?
So I ask once again-what energy do you think is going to leave the gas when that gas is used to force the piston down?
Yes I understand 'equal and opposite reactions' perfectly well.
If the gas is pushing a piston AND it moves then there is net force in the direction of travel of the cylinder. A force multiplied by distance travelled is work.
The amount of heat equal to the integral of F.ds LEAVES the gas.
The mechanism by which this happens is as I have described before.
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It is you that is clueless.
Im refering to the gas pushing the piston down, not the piston compressing the gas. This brings the point that you simply dont understand equal and opposites. If the gas is pushing on the piston, then the piston is also pushing on the gas just as hard. So as the pressure bjilds, then so dose the heat. So, the heavier the load, the faster the pressure builds.
Do you not understand equal and opposite reactions LE?
So I ask once again-what energy do you think is going to leave the gas when that gas is used to force the piston down?
Tinman, LE is hardly clueless. He has covered both the compression and expansion cases. There are only three general possibilities: What we are telling you is wrong, what you are saying is wrong, or we are all wrong. Would you do me a favor and respond to my post #205? I want to make sure that we see eye to eye on what it is that you believe.
The gas is always pushing in all directions. How much it pushes depends on the kinetic energy in the gas: How many molecules are bouncing around and how fast they are moving. The temperature is a proxy for how fast they move. What LE tried to explain in his last post was that the process of changing the volume IE the action of moving the piston either adds energy to the molecules, or removes it. The internal energy, commonly referred to as the heat changes up or down, and since the number of molecules stays the same, the temperature goes up and down with the heat. This principle of external mechanical work exchange with internal energy of a gas is both a very fundamental and key concept in thermodynamics. I know you don't believe this. Rather than follow a useless "Yes it is. No it isn't." exchange, I would like to help you understand that this is really true.
Imagine that you have an elastic balloon filled with ordinary dry air. Would you agree that the pressure inside the balloon is balanced by the spring force of the stretched balloon, and the outside air pressure? Now, if we put that balloon in an environmental chamber and pump the chamber down to 0.5 atmospheres, would you expect the balloon to expand? Would you agree that there is now more energy stored in the balloon membrane than before? Would you also agree that there is less energy stored in the external atmosphere? Can I then persuade you that since the balloon body energy has gone up, and the atmosphere energy has gone down that the gas inside the balloon had to give up energy? If you don't find this persuasive, then please explain where you think the balloon body got the extra energy that is now stored in its further stretched membrane.
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Yes I understand 'equal and opposite reactions' perfectly well.
If the gas is pushing a piston AND it moves then there is net force in the direction of travel of the cylinder. A force multiplied by distance travelled is work.
The amount of heat equal to the integral of F.ds LEAVES the gas.
The mechanism by which this happens is as I have described before.
That is where you are wrong, and if you understood the workings of the diagram on the first page then you would see where your mistake is being made. No heat leaves the gas as the piston is moving, as the pressure is continually rising within the cylinder-regardless of cylinder volume. The very same thing is happening as it is if you compress the gas within the cylinder via pushing the piston in and reducing cylinder volume will maintaining gas mass amount.
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That is where you are wrong, and if you understood the workings of the diagram on the first page then you would see where your mistake is being made. No heat leaves the gas as the piston is moving, as the pressure is continually rising within the cylinder-regardless of cylinder volume. The very same thing is happening as it is if you compress the gas within the cylinder via pushing the piston in and reducing cylinder volume will maintaining gas mass amount.
Irrespective of your diagram (which is wrong in terms of energy balance), heat leaves the cylinder when it does work on the environment and vice versa. Where does it leave to? It gets converted to/from work.
if it didn't, you would have already invented a free energy device and would have no need to add the extra complications of electrolysis etc.
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That is where you are wrong, and if you understood the workings of the diagram on the first page then you would see where your mistake is being made. No heat leaves the gas as the piston is moving, as the pressure is continually rising within the cylinder-regardless of cylinder volume. The very same thing is happening as it is if you compress the gas within the cylinder via pushing the piston in and reducing cylinder volume will maintaining gas mass amount.
Let's look at that idea:
In your earlier description you open valve 'A' after the vessel reaches 120psi. The piston then moves against the spring. Suppose the gas is evolving at 1 liter per minute and the cylinder volume can expand by 1 liter. Do you agree that if the piston extends fully in less than one minute that the overall pressure will have dropped from the time the valve is opened to the time the piston fully extends?
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Tinman, LE is hardly clueless. He has covered both the compression and expansion cases. There are only three general possibilities: What we are telling you is wrong, what you are saying is wrong, or we are all wrong. Would you do me a favor and respond to my post #205? I want to make sure that we see eye to eye on what it is that you believe.
The gas is always pushing in all directions. How much it pushes depends on the kinetic energy in the gas: How many molecules are bouncing around and how fast they are moving. The temperature is a proxy for how fast they move. What LE tried to explain in his last post was that the process of changing the volume IE the action of moving the piston either adds energy to the molecules, or removes it. The internal energy, commonly referred to as the heat changes up or down, and since the number of molecules stays the same, the temperature goes up and down with the heat. This principle of external mechanical work exchange with internal energy of a gas is both a very fundamental and key concept in thermodynamics. I know you don't believe this. Rather than follow a useless "Yes it is. No it isn't." exchange, I would like to help you understand that this is really true.
Imagine that you have an elastic balloon filled with ordinary dry air. Would you agree that the pressure inside the balloon is balanced by the spring force of the stretched balloon, and the outside air pressure? Now, if we put that balloon in an environmental chamber and pump the chamber down to 0.5 atmospheres, would you expect the balloon to expand? Would you agree that there is now more energy stored in the balloon membrane than before? Would you also agree that there is less energy stored in the external atmosphere? Can I then persuade you that since the balloon body energy has gone up, and the atmosphere energy has gone down that the gas inside the balloon had to give up energy? If you don't find this persuasive, then please explain where you think the balloon body got the extra energy that is now stored in its further stretched membrane.
No-the energy within the gas in the ballon drops when the outside pressure around the baloon drops, as the pressure of the gas within the baloon drops as the size of the vessel (baloon) increases.
You are assuming that either the pressure or temperature within the cylinder drops as the piston is pushed out by the gas-it dose not. When the piston is pushed out by the gas, the temperature, pressure, volume and gas mass increases. The heavier the force working against the pistons direction, the quicker the temperature and pressure will rise. The end result at the completion of the pistons travel against the aposing force is a higher temperature than that of the gas that was used to force the piston out in the first place.
This we see all the time on the air rams at the mine sites I have worked on. The air feeding the rams may be around the 25*C mark, while the air exiting the rams may be around 40*C. The heavier the load acting against the rams, the hotter the air leaving the rams.
There is no energy loss within my system when the gas is used to do outside work, as there is no temperature drop-but a temperature increase in the gas insted. This increase in temperature is an increase in energy-not a loss.
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No-the energy within the gas in the ballon drops when the outside pressure around the baloon drops, as the pressure of the gas within the baloon drops as the size of the vessel (baloon) increases.
So we are agreed that dropping the pressure outside the balloon causes the balloon to expand, and the gas within the balloon to exhange energy with the balloon membrane?
You are assuming that either the pressure or temperature within the cylinder drops as the piston is pushed out by the gas-it dose not. When the piston is pushed out by the gas, the temperature, pressure, volume and gas mass increases.
Whoa, the gas quantity (and therefore for particular gasses the gas mass) is changing slowly via the electrolysis. But we can open the valve as fast as we want. As soon as we do, the pressure in the cylinder before having been less than that of the gas in the pressure vessel, gas enters the cylinder, reducing pressure in the vessel, yes? The heavier the force working against the pistons direction, the quicker the temperature and pressure will rise.
I tell you what: Take an aerosol can, such as dust spray from the stationary store and let some of the gas out. Does the can get hotter or colder in your hand?The end result at the completion of the pistons travel against the aposing force is a higher temperature than that of the gas that was used to force the piston out in the first place.
Well according to classical thermodynamics, the gas that expands gets colder, that's the large volume of gas in the pressure vessel, while the gas in the cylinder gets hotter as it compresses.This we see all the time on the air rams at the mine sites I have worked on. The air feeding the rams may be around the 25*C mark, while the air exiting the rams may be around 40*C. The heavier the load acting against the rams, the hotter the air leaving the rams.
Again: compressing gas does work on it, and it gets hotter, expanding gas gets colder. The tanks feeding your rams cool down more and more the greater the volume of gas that is used per ram cycle.There is no energy loss within my system when the gas is used to do outside work, as there is no temperature drop-but a temperature increase in the gas insted. This increase in temperature is an increase in energy-not a loss.
Well as LE said: Woo hoo, that's free energy. Except that it isn't. All of the mechanical work comes at the cost of gas in the pressure vessel and cylinder. The cylinder heats up at the start of the stroke when the compressed gas is admitted by opening the valve and before the piston starts moving. As the piston moves, that gas expands and cools. In the meantime the gas in the pressure vessel has also expanded and cooled and continues to do so throughout the cylinder stroke.
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So we are agreed that dropping the pressure outside the balloon causes the balloon to expand, and the gas within the balloon to exhange energy with the balloon membrane?Whoa, the gas quantity (and therefore for particular gasses the gas mass) is changing slowly via the electrolysis. But we can open the valve as fast as we want. As soon as we do, the pressure in the cylinder before having been less than that of the gas in the pressure vessel, gas enters the cylinder, reducing pressure in the vessel, yes?I tell you what: Take an aerosol can, such as dust spray from the stationary store and let some of the gas out. Does the can get hotter or colder in your hand?Well according to classical thermodynamics, the gas that expands gets colder, that's the large volume of gas in the pressure vessel, while the gas in the cylinder gets hotter as it compresses.Again: compressing gas does work on it, and it gets hotter, expanding gas gets colder. The tanks feeding your rams cool down more and more the greater the volume of gas that is used per ram cycle.Well as LE said: Woo hoo, that's free energy. Except that it isn't. All of the mechanical work comes at the cost of gas in the pressure vessel and cylinder. The cylinder heats up at the start of the stroke when the compressed gas is admitted by opening the valve and before the piston starts moving. As the piston moves, that gas expands and cools. In the meantime the gas in the pressure vessel has also expanded and cooled and continues to do so throughout the cylinder stroke.
Yes, the gas in the pressure vessel cool as the ram expands.
No-the gas in the ram dosnt cool as it expands as the mass of gas is increasing, along with pressure. The spring requires a progresive rise in pressure to compress the spring-this is regardless of any other external force acting upon the piston. So while there is a loss at the pressure vessel, there is an equal gain in the cylinder.
Once again-the temperature, pressure, volume and gas mass rises in the ram progressively.
If I take a 10 ltr tank, and fill that tank with air to a pressure of 50psi, I can make that ram perform external work without loosing any energy that is avaliable to do work within that stored tank of air-that tank of air can do the same work with or without the ram opperating.
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Air ram and valves now ordered.
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Yes, the gas in the pressure vessel cool as the ram expands.
No-the gas in the ram dosnt cool as it expands as the mass of gas is increasing, along with pressure. The spring requires a progresive rise in pressure to compress the spring-this is regardless of any other external force acting upon the piston. So while there is a loss at the pressure vessel, there is an equal gain in the cylinder.
Once again-the temperature, pressure, volume and gas mass rises in the ram progressively.
Yes the ram heats and the pressure vessel cools, we agree on that. Where we do not agree is the idea that you do not lose energy from the combined gas volumes between the time you open the cylinder valve, and the ram reaches its equilibrium extension which is basic thermodynamics.If I take a 10 ltr tank, and fill that tank with air to a pressure of 50psi, I can make that ram perform external work without loosing any energy that is avaliable to do work within that stored tank of air-that tank of air can do the same work with or without the ram opperating.
I know you believe that despite the fact that it is dead wrong. The best thing for you to do is conduct your experiments.
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Yes the ram heats and the pressure vessel cools, we agree on that. Where we do not agree is the idea that you do not lose energy from the combined gas volumes between the time you open the cylinder valve, and the ram reaches its equilibrium extension which is basic thermodynamics.
No-its not basic thermodynamics,it's basic physics. For every action there is an equal and opposite reaction=what you loose in one,you gain in another. The other fact that supports this is-->you simply cannot loose(destroy)energy. Heat energy being disipated into the enviroment is not looseing energy,it is transfering energy. Well insulated pipes and tanks will see this transfer brought down to a minimum.
I know you believe that despite the fact that it is dead wrong. .
No-it is not dead wrong. When the gas is used to apply a force to a resistance,the resistance applies the very same amount of force on the gas-->equal and opposites.
You also assume (incorrectly)that the gas cools as it leaves the pressure vessel-->it dose not. You asked why the outlet of a pressure vessel gets cold when the gas is flowing out through that outlet,well that is because the flowing gas is drawing out the stored heat within that outlet nozzel,and thus,the nozzel freezes. This can only mean that the gas leaving the pressure vessel is hotter than that stored within the pressure vessel,but misconseptions come in regards to the temperature of that gas,as it mixes with the ambiant air once it leaves the pressure vessel.
The best thing for you to do is conduct your experiments.
And what makes you think i havnt already ?.
Im simply upgrading all my equipment,so as a clear and precise result can be shown. Everything will be automated-->no hands will be used that could be considered to be contributing to the end result's.
Cost of this project will be around the $500.00 mark. I simply would not spend this type of money on something i hadnt already tried.
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No-its not basic thermodynamics,it's basic physics. For every action there is an equal and opposite reaction=what you loose in one,you gain in another.
You are misapplying Newton's Third Law. The other fact that supports this is-->you simply cannot loose(destroy)energy.
That's right you can't destroy it and you cannot create it either. When you convey energy by doing work on the external system, you lose that energy from the internal system. Heat energy being disipated into the enviroment is not looseing energy,it is transfering energy.
That's right. Well insulated pipes and tanks will see this transfer brought down to a minimum.
All the insulation in the world is not going to prevent the internal energy loss that results from performing work on the external system.
No-it is not dead wrong. When the gas is used to apply a force to a resistance,the resistance applies the very same amount of force on the gas-->equal and opposites.
If the forces are equal then the system is in equilibrium. When the forces are unequal then Newton's Second and Third Laws both come into play: The system accelerates according to the difference in force divided by the mass.You also assume (incorrectly)that the gas cools as it leaves the pressure vessel-->it dose not.
But it does. This is really basic thermodynamics. Do your experiments, show your measurements and then we can talk. Boyle's Law which is a subset of the ideal gas law applies here. You don't believe that. We have covered it many times. A game of: "Yes it is. No it isn't." is not productive. You asked why the outlet of a pressure vessel gets cold when the gas is flowing out through that outlet,well that is because the flowing gas is drawing out the stored heat within that outlet nozzel,and thus,the nozzel freezes.
The reason that a vessel of compressed gas cools as gas is released is due to Boyle's Law. This can only mean that the gas leaving the pressure vessel is hotter than that stored within the pressure vessel,but misconseptions come in regards to the temperature of that gas,as it mixes with the ambiant air once it leaves the pressure vessel.
The vessel cools as the gas is released. We can start with the vessel in absolute thermal equilibrium with the outside atmosphere and this will always be true. I know you don't believe this, despite what you can find out with a $3. can of dusting spray and a your hand as a temperature sensor. So. do your experiments.
And what makes you think i havnt already ?.
I make no assumptions about what you have or have not done. When you report your data we can talk about how you performed your experiments, the measurement methods and the data. You haven't reported those things so we are not in a position to talk about them.Im simply upgrading all my equipment,so as a clear and precise result can be shown. Everything will be automated-->no hands will be used that could be considered to be contributing to the end result's.
I for one appreciate that you put time and effort into your experiments. I think that you really want to learn the truth.Cost of this project will be around the $500.00 mark. I simply would not spend this type of money on something i hadnt already tried.
You have your expectations. That is all fine and well. We will see how you go about your tests, and what data you end up with in the end.
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When you convey energy by doing work on the external system, you lose that energy from the internal system
I ask this question in all seriousness Mark, as I do appreciate your stringent adherence to real proof as you define it. I do disagree with you there on a couple of points.
Although I really do appreciate you not taking things personal even when provoked (with some exceptions)
As a complete novice here is something I don't understand that I really would appreciate your position on the Big Bang theory ( not the tv show) the situation you describe above does not seem to fit.
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I ask this question in all seriousness Mark, as I do appreciate your stringent adherence to real proof as you define it. I do disagree with you there on a couple of points.
Although I really do appreciate you not taking things personal even when provoked (with some exceptions)
As a complete novice here is something I don't understand that I really would appreciate your position on the Big Bang theory ( not the tv show) the situation you describe above does not seem to fit.
I think that the Bernadette and Amy characters are masterful. Oh right, not the TV show.
There are some things on which we presently have a very weak or no grasp. One of them is what was there before there was anything. The strong evidence for the Big Bang is the microwave background. What caused there to be a primordial hot stew of stuff and what caused it to spew forth and start cooling is not something for which I have seen a satisfying explanation. The very rude guy who likes to say that a bunch of great minds were full of hooey may be somewhat right about that. The trouble is that greater minds or at least minds with better insights seem to be lacking. QM is very strange stuff, so much so that it's easy to get the idea that it's analagous to Ptolemy's epicycles all over again: A set of rules that in a very complicated way makes accurate predictions within its limits. When it comes to QED the predictions are really, really good. That still doesn't mean that QED is the right idea, but it sure makes it very practical to use.
Maybe someone will come along with something far more elegant that ties to a more fundamental understanding of what makes the universe tick as opposed to describing the way that things about it appear to tick. If we don't get that, it would at least be nice to discover something that makes QM and QED less weird. In the interim, we muddle on.
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Mmmm thanks for taking the time to respond. btw I frigging hate the tv show.
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That's right you can't destroy it and you cannot create it either. When you convey energy by doing work on the external system, you lose that energy from the internal system.
You loose that energy to where Mark?. A reminder that the heading of this thread is !!open systems!!. The energy is not lost,and regardless wether or not there is a resistance against the ram,the energy state of the gas within that ram will not change,in fact,it gains energy in the way of heat when the ram is made to do work.
That's right.All the insulation in the world is not going to prevent the internal energy loss that results from performing work on the external system.
There is no internal loss when the gas is performing external work,there is a gain in heat energy within the cylinder(the gas) when that ram is made to do work.
If the forces are equal then the system is in equilibrium. When the forces are unequal then Newton's Second and Third Laws both come into play:
Every action has an equal and opposite reaction-->the forces are always equal in opposite directions. You cannot have two apposing forces that are unequal,otherwise perpetual motion would be possable. When you have what you call two unequal forces,you still only have a net force in one direction. This net force that is used to accelerate a mass is in the opposite direction to and equal to the accelerated mass-thus there is no unequal forces.
The reason that a vessel of compressed gas cools as gas is released is due to Boyle's Law.The vessel cools as the gas is released. We can start with the vessel in absolute thermal equilibrium with the outside atmosphere and this will always be true. I know you don't believe this, despite what you can find out with a $3. can of dusting spray and a your hand as a temperature sensor.
I have never said that i dont believe the vessel will cool when the gas is released,in fact,i have said the opposite-->so not sure where you came up with that statement. What i am saying is that the gas temperature in the ram will increase when the ram is made to do work-external work. So while the energy within the vessel reduces,the energy within the ram increases by the same amount-no energy is lost within the closed system. The energy stored at this time(after the ram has performed work on the external system) within both the vessel and ram is the very same amount as we started with within the vessel.
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You loose that energy to where Mark?. A reminder that the heading of this thread is !!open systems!!. The energy is not lost,and regardless wether or not there is a resistance against the ram,the energy state of the gas within that ram will not change,in fact,it gains energy in the way of heat when the ram is made to do work.
Energy from the internal system is transferred to the external system. The internal system no longer has the transferred energy. In your arrangement: the average gas temperature goes down. The sequential flow of energy is: vessel gas volume => ram gas volume => external load. In each of the two transfers the entropy of the system increases. IE you end up with more mass at a lower average temperature at the end of each of the two steps than at the beginning of each.
There is no internal loss when the gas is performing external work,there is a gain in heat energy within the cylinder(the gas) when that ram is made to do work.
Round and round we go with no progress. Do your experiments or if you have already done them, publish them and we can talk about what you see.Every action has an equal and opposite reaction-->the forces are always equal in opposite directions.
Newton's Third Law translates to: "Any acceleration of a mass in one direction (action) imposes an equal and opposing acceleration of mass (reaction) in the opposite direction." Because Newton's Second Law equates force to mass multiplied by acceleration, Newton's Third Law also translates to: "Any force or mass acceleration has an equal and opposite force or mass acceleration." When one or the other of the masses is very large compared to the other, the smaller mass will accelerate at a far greater rate than the larger mass. You cannot have two apposing forces that are unequal,otherwise perpetual motion would be possable.
No, that is not what Newton's Third Law means. N3 does not address energy. N3 addresses force. If we had an infinite supply of energy and propellant, N3 means that we could travel the stars forever. That is in no way perpetual motion. When you have what you call two unequal forces,you still only have a net force in one direction. This net force that is used to accelerate a mass is in the opposite direction to and equal to the accelerated mass-thus there is no unequal forces.
If you don't have a static force to oppose another, acceleration results, and by Newton's Second Law, action and reaction are then matched.
I have never said that i dont believe the vessel will cool when the gas is released,in fact,i have said the opposite-->so not sure where you came up with that statement. What i am saying is that the gas temperature in the ram will increase when the ram is made to do work-external work.
The temperature in the ram must increase before the ram can do work. In the course of doing work, the temperature in the ram decreases from that elevated level. Again: energy flows from the vessel gas cooling the vessel gas, and heating the ram gas, and then from the ram gas into the external load cooling the ram gas. Energy for any of the gas volumes at any point is: n*R*T. n does not change, and R is a constant. As energy is transferred from one gas volume to someplace else the temperature of that gas volume with its fixed number of molecules n decreases. So while the energy within the vessel reduces,the energy within the ram increases by the same amount-no energy is lost within the closed system.
If the ram could not do work on the external system, and there were no heat transfer to the outside system then that would be true. But such is not the case, because in order for any gas to pass from the pressure vessel to the ram, it must pass through the valve orifice and in doing so will lose energy to the orifice body. If the container were perfectly insulated from the outside world then eventually the orifice would reheat the gas. The energy stored at this time(after the ram has performed work on the external system) within both the vessel and ram is the very same amount as we started with within the vessel.
You keep saying this. We keep going round and round without progress. Let's see what your experiments seem to show.
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Mmmm thanks for taking the time to respond. btw I frigging hate the tv show.
Tastes vary. The show is very popular. I like it. I think that the Bernadette and Amy characters are cleverly written. I think the mothers they cooked up for Sheldon and Leonard are priceless.
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...the energy within the ram increases by the same amount-no energy is lost within the closed system. The energy stored at this time(after the ram has performed work on the external system) within both the vessel and ram is the very same amount as we started with within the vessel...
etc. etc.
Absolutely amazing! Using this incredible theoretical breakthrough I have been able to design a brand new ICE engine. It only requires a little bit of petrol to start the cycle and then it runs forever!.. The oil industry is doomed..!
Sorry for sounding patronizing but you have absolutely zero clue about thermodynamics. When a gas expands and does external work heat is lost from the gas. Until you can understand why that is you have no hope of understanding why you device can never work.
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Absolutely amazing! Using this incredible theoretical breakthrough I have been able to design a brand new ICE engine. It only requires a little bit of petrol to start the cycle and then it runs forever!.. The oil industry is doomed..!
Sorry for sounding patronizing but you have absolutely zero clue about thermodynamics. When a gas expands and does external work heat is lost from the gas. Until you can understand why that is you have no hope of understanding why you device can never work.
This has been a long and frustrating journey for all. I think that tinman is honest in his intentions. If it is at all possible, I would like to carry this through to where he can learn the truth from his experiments. With that in mind, I ask for additional patience with tinman.
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Tastes vary. The show is very popular. I like it. I think that the Bernadette and Amy characters are cleverly written. I think the mothers they cooked up for Sheldon and Leonard are priceless.
The only episode of that show I really liked was where Woz made a guest appearance.
Bill
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Absolutely amazing! Using this incredible theoretical breakthrough I have been able to design a brand new ICE engine. It only requires a little bit of petrol to start the cycle and then it runs forever!.. The oil industry is doomed..!
SARCASTIC CLOWN.
Sorry for sounding patronizing but you have absolutely zero clue about thermodynamics. When a gas expands and does external work heat is lost from the gas. Until you can understand why that is you have no hope of understanding why you device can never work.
Until you understand the whole workings of the completed device,insted of jumping in with rubbish half way through showing one part of the device,you should keep you inaccurate bable to your self.
This is part of a bouyancy/gravity devcice,and the ram is to show the expanding bladder that will be part of this device as a whole. So Einstein,let me ask you this. If we take a balloon and fill it with air,will it have a greater upward force when at 20 meters below the sea when the gas is more compressed inside the balloon,or will it have more upward force when at 5 meters below the sea where the gas inside the balloon is less compressed?. So at 20 meters below sea level,the balloon may have a displacement area of say(hypothetical)1 cubic foot,and at 5 meters it may have a displacement area of say(hypothetical)1.5 cubic feet. So which now has the potential of producing the greater amount of energy Einstein-->the expanded gas,or the more compressed gas?.
Until you understand the complete workings of the device,please keep your idiotic insults to your self.
TinMan.
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This has been a long and frustrating journey for all. I think that tinman is honest in his intentions. If it is at all possible, I would like to carry this through to where he can learn the truth from his experiments. With that in mind, I ask for additional patience with tinman.
Please make sure that you read my last comment to Mr Einstein Mark,as i think you are thinking along the wrong lines here. You may need to go back to the start,and refresh on the fact that this device as a whole is a bouyancy/gravity device. The ram is to simulate an expanding bouyant device,but in this first test,it will have a resistive force applied to it to show that an external force dose not change the end result of the ram being filled with gas from the vessel. If the vessel supplies the ram with the gas required to reach say 50psi inside that ram when fully extended,then making that ram do work will not result in a further reduction of energy within the gas.-->remember at the start the gas that was to be used was HHO.
Now,i need to know if you can convert a given volume of gas at a set pressure and temerature to joules of energy?. This will be needed when calculating total energy at the start of the test,and total energy at the end of the test.
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SARCASTIC CLOWN.
Until you understand the whole workings of the completed device,insted of jumping in with rubbish half way through showing one part of the device,you should keep you inaccurate bable to your self.
This is part of a bouyancy/gravity devcice,and the ram is to show the expanding bladder that will be part of this device as a whole. So Einstein,let me ask you this. If we take a balloon and fill it with air,will it have a greater upward force when at 20 meters below the sea when the gas is more compressed inside the balloon,or will it have more upward force when at 5 meters below the sea where the gas inside the balloon is less compressed?. So at 20 meters below sea level,the balloon may have a displacement area of say(hypothetical)1 cubic foot,and at 5 meters it may have a displacement area of say(hypothetical)1.5 cubic feet. So which now has the potential of producing the greater amount of energy Einstein-->the expanded gas,or the more compressed gas?.
Until you understand the complete workings of the device,please keep your idiotic insults to your self.
TinMan.
At some stage in the future you will realise you have made a fundamental error in your understanding of the nature of the conversion of heat to work and vice versa.
I'm only actually trying to help you from wasting your time on this unworkable concept.
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At some stage in the future you will realise you have made a fundamental error in your understanding of the nature of the conversion of heat to work and vice versa.
I'm only actually trying to help you from wasting your time on this unworkable concept.
There is no loss or gain in a closed system,the gain comes from the open half of the system,which in turn come from nature. If we adopt your attitude LE,we will go no further than we already have. Man puts in X amount of energy to make a solar pannel,and nature returns this X amout of energy-->plus a whole lot more.The closed system in this case is the solar pannel/charge controller/battery/load. The open part of the system is the sun. Is the sun effected by any load placed on the solar pannels output? NO,and neither is the gases energy effected by an external resistance placed apon the ram-the open part of the system as a whole.
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There is no loss or gain in a closed system,the gain comes from the open half of the system,which in turn come from nature. If we adopt your attitude LE,we will go no further than we already have. Man puts in X amount of energy to make a solar pannel,and nature returns this X amout of energy-->plus a whole lot more.The closed system in this case is the solar pannel/charge controller/battery/load. The open part of the system is the sun. Is the sun effected by any load placed on the solar pannels output? NO,and neither is the gases energy effected by an external resistance placed apon the ram-the open part of the system as a whole.
Funny you should mention the sun. I have no link to this but could probably find it...a "study" (done by global warming believer liberals) cautions against adopting solar panels world wide as this would "stress" the sun and cause it to burn out much earlier than it normally would.
I kid you not. Now I, of course, do not believe this for a minute but, someone, or some people got funded by the US Gov. and actually came up with this crap!! This reminds me of the "study" that warned against the use of hydrogen powered cars as it would deplete the world's water supply. My question would be...where is the water going? Into space?
Anyway, just wanted to share this stuff.
Bill
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There is no loss or gain in a closed system,the gain comes from the open half of the system,which in turn come from nature.
First, you use the term 'closed system' to describe what is actually referred to as an 'isolated system'. i.e. a boundary over which neither matter NOR energy may pass. In actual fact isolated systems do not occur in nature but as engineers we sometimes pretend they do.
A closed system may pass energy , but not matter, and an 'open system' may pass both across a boundary.
These boundaries are completely arbitrary and are only chosen for the convenience of analysis. If you account for heat and/or work flows across them then it matters not where they are drawn.
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Please make sure that you read my last comment to Mr Einstein Mark,as i think you are thinking along the wrong lines here. You may need to go back to the start,and refresh on the fact that this device as a whole is a bouyancy/gravity device.
Well, based on the descriptions to date, the buoyancy angle is a new one on me. Be careful that you haven't mixed so many different things together that you cannot separate the individual effects. The ram is to simulate an expanding bouyant device,but in this first test,it will have a resistive force applied to it to show that an external force dose not change the end result of the ram being filled with gas from the vessel. If the vessel supplies the ram with the gas required to reach say 50psi inside that ram when fully extended,then making that ram do work will not result in a further reduction of energy within the gas.-->remember at the start the gas that was to be used was HHO.
Yes I recall that you said you were electrolyzing water to generate your gas. If you change that, just say what it is you are doing.
Now,i need to know if you can convert a given volume of gas at a set pressure and temerature to joules of energy?.
If the gas is known, then yes that is readily done. What will you be using for temperature and pressure measurements? This will be needed when calculating total energy at the start of the test,and total energy at the end of the test.
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There is no loss or gain in a closed system,the gain comes from the open half of the system,which in turn come from nature. If we adopt your attitude LE,we will go no further than we already have. Man puts in X amount of energy to make a solar pannel,and nature returns this X amout of energy-->plus a whole lot more.The closed system in this case is the solar pannel/charge controller/battery/load. The open part of the system is the sun. Is the sun effected by any load placed on the solar pannels output? NO,and neither is the gases energy effected by an external resistance placed apon the ram-the open part of the system as a whole.
Well, actually: The sun emits energy and barring any signficant reflections, that energy is gone. When a volume of gas expands by driving a piston, the machine does work and loses the internal energy delivered as work to the outside world. Then that energy is gone as well. I know that you don't believe this, but if you stick with your experiments you should be able to learn it by the time you are done.
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Funny you should mention the sun. I have no link to this but could probably find it...a "study" (done by global warming believer liberals) cautions against adopting solar panels world wide as this would "stress" the sun and cause it to burn out much earlier than it normally would.
What lunatics propose such an idea???
I kid you not. Now I, of course, do not believe this for a minute but, someone, or some people got funded by the US Gov. and actually came up with this crap!! This reminds me of the "study" that warned against the use of hydrogen powered cars as it would deplete the world's water supply. My question would be...where is the water going? Into space?
Anyway, just wanted to share this stuff.
Bill
Stupid ideas are not hard to find.
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What lunatics propose such an idea???Stupid ideas are not hard to find.
This was a few months ago on the top of the hour news on my local radio station. (CBS radio news) I mean the one about the sun being drained. The other report I do not recall where that came from as it was about a year ago or so. I can look for it..it just pisses me off that we taxpayers foot the bill for these stupid studies done by God knows who.
Bill
http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/ (http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/)
Above is a link to the sun depletion study.
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This was a few months ago on the top of the hour news on my local radio station. (CBS radio news) I mean the one about the sun being drained. The other report I do not recall where that came from as it was about a year ago or so. I can look for it..it just pisses me off that we taxpayers foot the bill for these stupid studies done by God knows who.
Bill
http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/ (http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/)
Above is a link to the sun depletion study.
It reads like parody to me.
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Funny you should mention the sun. I have no link to this but could probably find it...a "study" (done by global warming believer liberals) cautions against adopting solar panels world wide as this would "stress" the sun and cause it to burn out much earlier than it normally would.
I kid you not. Now I, of course, do not believe this for a minute but, someone, or some people got funded by the US Gov. and actually came up with this crap!! This reminds me of the "study" that warned against the use of hydrogen powered cars as it would deplete the world's water supply. My question would be...where is the water going? Into space?
Anyway, just wanted to share this stuff.
Bill
You can write them a letter telling them that this is not true. The sun radiates energy that is lost anyways (Lost from the suns perspective). If someone harvest some of this energy and convert it into work, it will definitely not have any influence on the sun.
IF the sun was influenced by this, think of how much energy the sun radiates. Here on earth that is approx 1kW pr. square meter. Multiply that with how many square meters the sphere with a diameter equal to 2 x radius between the earth and the sun has. How many watt do you get, and how many watt do you "drain" from the sun? The "drain" is in any case negligible.
The sun radiates approx 2 x1027 Joule pr. second. If we "drain" 1GJ pr. second from this energy for the rest of the suns lifetime, the sun would not last for 5 billion more years, but 79 milliseconds sooner.
If the half the surface of the earth (Well, actually the area of the earth cross section) is used to harvest all 1kW pr square meters, the lifetime of the sun would be reduced by 112 days...
If you drain water out of a bottle, and somewhere below the bottle put a turbine or some impellers, it will not influence how fast the bottle is drained.
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You can write them a letter telling them that this is not true. The sun radiates energy that is lost anyways (Lost from the suns perspective). If someone harvest some of this energy and convert it into work, it will definitely not have any influence on the sun.
IF the sun was influenced by this, think of how much energy the sun radiates. Here on earth that is approx 1kW pr. square meter. Multiply that with how many square meters the sphere with a diameter equal to 2 x radius between the earth and the sun has. How many watt do you get, and how many watt do you "drain" from the sun? The "drain" is in any case negligible.
The sun radiates approx 2 x1027 Joule pr. second. If we "drain" 1GJ pr. second from this energy for the rest of the suns lifetime, the sun would not last for 5 billion more years, but 79 milliseconds sooner.
If the half the surface of the earth (Well, actually the area of the earth cross section) is used to harvest all 1kW pr square meters, the lifetime of the sun would be reduced by 112 days...
If you drain water out of a bottle, and somewhere below the bottle put a turbine or some impellers, it will not influence how fast the bottle is drained.
Now we are starting to see what open systems are.
We can replicate the sun experiment at home. If we take say a 5 watt 12 volt incandescent bulb,that 5 watts of power is transformed into light and heat energy. If we now place solar pannels around that light bulb,and place a load on the output from those solar pannels,the bulb will still emit the same amount of light and heat for that 5 watts of power-->the solar pannels place no load on the closed system(the light bulb and battery). So all of that 5 watts is accounted for by way of heat and light emissions from the bulb,and we can now draw power from the solar pannels without cost or reflection on the closed system.
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Now we are starting to see what open systems are.
We can replicate the sun experiment at home. If we take say a 5 watt 12 volt incandescent bulb,that 5 watts of power is transformed into light and heat energy. If we now place solar pannels around that light bulb,and place a load on the output from those solar pannels,the bulb will still emit the same amount of light and heat for that 5 watts of power-->the solar pannels place no load on the closed system(the light bulb and battery). So all of that 5 watts is accounted for by way of heat and light emissions from the bulb,and we can now draw power from the solar pannels without cost or reflection on the closed system.
LE's definition of open and closed is correct. Anyway the silly item PB found that I am quite certain is a parody brings up a point that we see in free energy claims a lot: The more decoupled a source and a load appear to be, the less efficient energy transfer tends to be. The incident radiation from the sun on the earth happens pretty much no matter what. Even if the earth were wrapped in a perfect mirror, the return radiation to the sun would be insignificant versus the total solar flux.
In the machine that you have described so far, if the piston drives an external load, then the piston is tightly coupled to the gas volume in the cylinder.
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This was a few months ago on the top of the hour news on my local radio station. (CBS radio news) I mean the one about the sun being drained. The other report I do not recall where that came from as it was about a year ago or so. I can look for it..it just pisses me off that we taxpayers foot the bill for these stupid studies done by God knows who.
Bill
http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/ (http://www.provedplusprobable.com/solar-power-discovered-to-be-draining-the-sun/)
Above is a link to the sun depletion study.
I remember this one,and another study by so called scientist that said current groth rate of solar pannel use would see the sun go out in 400 years lol.
It is interesting that both of the scientific studies were funded by petrochemical companies lol.
They have even given it a fancy name-->forced photovoltaic drainage.” ???
!!Apparently!! If every home in the world had solar panels on their roofs, global temperatures would drop by as much as thirty degrees over twenty years, and the sun could die out within three hundred to four hundred years.” LOL
http://www.energymatters.com.au/renewable-news/em4318/
Rest easy guy's,here is the whole story.
http://www.politifact.com/truth-o-meter/statements/2014/may/29/chain-email/claim-solar-panels-drain-suns-energy-are-satire-no/
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Now we are starting to see what open systems are.
We can replicate the sun experiment at home. If we take say a 5 watt 12 volt incandescent bulb,that 5 watts of power is transformed into light and heat energy. If we now place solar pannels around that light bulb,and place a load on the output from those solar pannels,the bulb will still emit the same amount of light and heat for that 5 watts of power-->the solar pannels place no load on the closed system(the light bulb and battery). So all of that 5 watts is accounted for by way of heat and light emissions from the bulb,and we can now draw power from the solar pannels without cost or reflection on the closed system.
In fact MarkE,could this not be seen as a system that puts out more energy than it consume's?
All of the P/in from the 12 volt battery(the 5 watts) is accounted for by way of both heat and light output from the bulb,and any power output from the solar pannels is excess. We know that loading the solar pannels will not reflect on the P/in to the light bulb,nor will it effect the light and heat output from the light bulb.
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In fact MarkE,could this not be seen as a system that puts out more energy than it consume's?
All of the P/in from the 12 volt battery(the 5 watts) is accounted for by way of both heat and light output from the bulb,and any power output from the solar pannels is excess. We know that loading the solar pannels will not reflect on the P/in to the light bulb,nor will it effect the light and heat output from the light bulb.
Why? Draw an imaginary boundary anywhere that you like and count the energy in and the energy out. Do you find that you can draw such a boundary and get more out than in? You do not get to count just the power coming out across some boundary. And you have to use opposite signs for energy crossing a boundary in one direction for energy crossing that same boundary in the opposite direction.
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Why? Draw an imaginary boundary anywhere that you like and count the energy in and the energy out. Do you find that you can draw such a boundary and get more out than in? You do not get to count just the power coming out across some boundary. And you have to use opposite signs for energy crossing a boundary in one direction for energy crossing that same boundary in the opposite direction.
Well i have no idea what that jibberish is with boundaries ???
It couldnt be simpler. Energy in is our 5 watts to the bulb,energy out has to be equal to that 5 watts-->unless you have a way to destroy energy?Our 5 watts of power flowing into the light bulb is converted to heat and light that equals the 5 watts-is it not?. And would the solar panels in any way change the amount of heat and light emmited from that bulb when a load is placed across those solar panels-->i think not. Would the load placed across those solap panel's reflect apon the energy source that is driving the light bulb-->i think not.
There you go,worlds first OU circuit-->couldnt be simpler lol ;D
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Well i have no idea what that jibberish is with boundaries ???
That could be a real problem. If you have trouble with boundaries then it is likely that you have problems accounting for energy, which you appear to have demonstrated above.
It couldnt be simpler. Energy in is our 5 watts to the bulb,energy out has to be equal to that 5 watts-->unless you have a way to destroy energy? Our 5 watts of power flowing into the light bulb is converted to heat and light that equals the 5 watts-is it not?.
That's correct, you draw an imaginary boundary around the light bulb, and the light bulb having negligible energy storage on average gets 5W in electrical and emits 5W out as heat and light. And would the solar panels in any way change the amount of heat and light emmited from that bulb when a load is placed across those solar panels-->i think not.
Of course not. So now you can either draw a boundary around the solar panels alone, and find that the energy again balances, or you can draw a boundary around both the light bulb and the solar panels. Or you can draw a boundary around the load connected to the solar panels, or the load and the solar panels, or the load, the solar panels, and the light bulb. In each case you should be able to account for all of the energy and find that what comes out is the difference of the input and what goes into or comes out of internal energy storage. Would the load placed across those solap panel's reflect apon the energy source that is driving the light bulb-->i think not.
Certainly not to any extent that we would bother trying to account.
There you go,worlds first OU circuit-->couldnt be simpler lol ;D
Perhaps such an idea might make it past the accountants at Enron. It would not make it past accountants much anywhere else.
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That could be a real problem. If you have trouble with boundaries then it is likely that you have problems accounting for energy, which you appear to have demonstrated above.
Lol,sounds like your boundaries are a mythical creature that you have invented. What you mean to say is-->we will continue to shift the goal post until such time that we make sure your device dose not exibit any sort of excess energy. This boundary you speak off would be between the two(loosly coupled) systems,as you call it. This boundary would also be between the one system that has no effect on the other,regardless of wether it's there in opperation or not(which also is the same boundary as the first).If you disagree with that Mark,then place your boundary,and explain as to why it's there in that position. If you cannot,then please refrain from continually stating that I have the problem with accounting for energy.
That's correct, you draw an imaginary boundary around the light bulb, and the light bulb having negligible energy storage on average gets 5W in electrical and emits 5W out as heat and light.
This boundary would be around system one as a whole-->the battery,conecting wires and globe is system one.This system is running at 100% efficiency when all energy in and out is accounted for.
So now you can either draw a boundary around the solar panels alone, and find that the energy again balances, or you can draw a boundary around both the light bulb and the solar panels.[/quote]
No. As i stated above,this boundary would be around each system as a whole. So now lets say our load on the solar panel is a 100 ohm resistor.
So boundary one is around system one(the battery,wires and light bulb)
The second boundary is around our open system(open from that of the first system),the solar panel and resistive load(100 ohm resistor). These two boundaries are set-->applying a load on,or removing the second system(system two)has no effect on system one.
Of course not.
Well whats the chances of that--> an accord :D
So there you have your boundaries Mark.
Boundary 1-is around system one. This system is running at 100% efficiency,where the energy in = the energy out. Net gain is 0.
Boundary 2-is around system 2(the solar panel,conecting wires and 100 ohm resistor)This boundary was set due to the fact that this system has no effect on system 1,regardless of wether it's there or not. This system is open to system one,and thus a sepperate boundary must be set around this system-->but im sure you would disagree with this!some how!
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Tinman:
In your example...5 watts into light bulb and 5 watts out in both heat and light...so far so good. Then you add a solar panel and want to count the electrical energy out of that panel claiming 0 input...is this correct?
If so, what I think is wrong here is that you are counting the 5 watts out in both heat and LIGHT output from the bulb. If some of that light is absorbed by the solar panel, and converted to electrical energy, then...by definition, you have "lost" some of the light output from the bulb. This loss would then be considered input to the the panel. I also think the panel would heat up a little thereby using more of your output from the bulb that you are not accounting for. That energy from the bulb (both heat and light) is going into the panel and therefore, you can't just say it is still there as it was before you added that panel. Some of that energy is now powering that solar panel and should be, must be, accounted for.
I could be wrong but, I don't think so.
Bill
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Lol,sounds like your boundaries are a mythical creature that you have invented. What you mean to say is-->we will continue to shift the goal post until such time that we make sure your device dose not exibit any sort of excess energy.
Pray tell what goal post has been moved? You complained that the notion of a system boundary did not make sense to you and I have as others before me have offered you an explanation.This boundary you speak off would be between the two(loosly coupled) systems,as you call it. This boundary would also be between the one system that has no effect on the other,regardless of wether it's there in opperation or not(which also is the same boundary as the first).
A boudnary is something that we draw for accounting reasons. You can draw the boundaries where you like. But once you do, the accounting rules are set. Those who do not aspire to reincarnate Enron do not get to add what crosses from outside a boundary to inside it along with what comes from inside the boundary to the outside using the same sign for each.If you disagree with that Mark,then place your boundary,and explain as to why it's there in that position. If you cannot,then please refrain from continually stating that I have the problem with accounting for energy.
This boundary would be around system one as a whole-->the battery,conecting wires and globe is system one.This system is running at 100% efficiency when all energy in and out is accounted for.
You have demonstrated double counting several times now. I leave it to you to define the boundaries anywhere you like. Once you do then we can all perform the accounting and contrast and compare.
Well whats the chances of that--> an accord :D
So there you have your boundaries Mark.
Boundary 1-is around system one. This system is running at 100% efficiency,where the energy in = the energy out. Net gain is 0.
Boundary 2-is around system 2(the solar panel,conecting wires and 100 ohm resistor)This boundary was set due to the fact that this system has no effect on system 1,regardless of wether it's there or not. This system is open to system one,and thus a sepperate boundary must be set around this system-->but im sure you would disagree with this!some how!
So, you have a light bulb: energy goes in and the same energy goes out. No excess. And you have a solar panel connected to a resistor. Again: the same amount of energy goes in as goes out. Where is your excess? Where is your anomaly? Next we come to your pressure vessel: Energy that goes out takes from what is inside. Ditto your ram. Ditto the combination.
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...Lol,sounds like your boundaries are a mythical creature that you have invented...
Of course they are and that is the whole point. They are boundaries of (usually) arbitrary dimensions and idealized thermodynamic properties that may not actually exist in nature but are chosen to represent the theoretical limits that any real system approaches but could never exceed.
Consider for instance the concept of the DALR or "dry adiabatic lapse rate". It is a very important concept in determining the stability of the atmosphere. It represents the theoretical rate that a parcel of dry air will cool at as it rises in altitude. As such it utilizes the concept of a closed thermodynamic system (in this case an adiabatic one) on an arbitrary parcel of air.
I use analysis based on the DALR it all the time by reading a Skew T Log P plot to determine if I should drive to the local sailplane airfield and try and set forth on a cross country flight. The predictions it provides by comparing the actual temperate plot with altitude compared with the behaviour of the fictional 'air parcel' provides a very accurate way of predicting what the days flying will be like.
Incidentally thermals do cool as they rise. They also expand (doing work) as they rise. If they didn't as per your faulty analysis I may have never bothered to come home. I'd still be up in the sky enjoying myself circuiting the earth effortlessly.
That theoretical tool is just one example of an analysis based on a 'mythical creature' but there are hundreds more.
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Tinman:
In your example...5 watts into light bulb and 5 watts out in both heat and light...so far so good. Then you add a solar panel and want to count the electrical energy out of that panel claiming 0 input...is this correct?
If so, what I think is wrong here is that you are counting the 5 watts out in both heat and LIGHT output from the bulb. If some of that light is absorbed by the solar panel, and converted to electrical energy, then...by definition, you have "lost" some of the light output from the bulb. This loss would then be considered input to the the panel. I also think the panel would heat up a little thereby using more of your output from the bulb that you are not accounting for. That energy from the bulb (both heat and light) is going into the panel and therefore, you can't just say it is still there as it was before you added that panel. Some of that energy is now powering that solar panel and should be, must be, accounted for.
I could be wrong but, I don't think so.
Bill
Hi Bill
I know what your saying,but think of it this way. The light bulb will emmit x amount of lumens of light regardless of wether the solar panel is there or not.The lumen output of the bulb will not change from the solar panel being open circuit,or has a load placed across it's output-->0 reflection shown from the solar panel in either case. The heat output is radiant,and that also dose not change regardless of the solar panel and its load. No matter how you look at it,the solar panel changes nothing in regards to the heat and light output of the bulb.The solar panel also will not show any reflection on the P/in to the light bulb.
Lets calculate and messure this system as we would any other.
First P/in-->this being a DC system is easy-->volts x amps.This is our energy consumption.
Then calculate the energy output from the bulb. Heat and light. This is also easy,as it is equal to the energy in-->energy can be neither created nor destroyed-only transformed.We also must take into account the heat from the conecting wires and battery it self. As this will be a radiant heat,then it is an energy output.
So we have accounted for all our energy in and out,and have a net result of 0.
Now system two-->our solar panel and resistive load(100 ohms in this case). This is also a DC output,so easy to calculate.
So now,we do as we always do. We place the load on the solar panel,and then remeassure the energy output from system 1. Do you think it will change when the solar panel(system 2)has a load across it? If system 1 shows no change when system 2 is put into opperation,then system 2 is generating excess energy,as the energy in system 1 has been accounted for in the way of heat and light.
So in summery
System 2 dosnt change the heat output of system 1
System 2 dosnt change the light output of system 1
And system 2 dosnt reflect on the P/in of system 1
If the solar panel was open(no load across it) then it would be said that it isnt consuming any energy,as energy cannot be destroyed.If nothing is comming out,then nothing is going in. So all you have to do now Bill,is place a load across that solar panel,and show me any change in either the energy in(watts) to system 1,or the energy out of system 1,and i'll believe what you say ;)
Whats happening here is a case of-->it's to simple to be true.
But if we had a transformer where the secondary shows no reflection on the primary what so ever,then every one would be jumping for joy. All we have here is that transformer,where light is converted into electrical energy,but shows no reflection on the primary.
MarkE will throw all sorts of curve balls at you(eg.boundaries)but never put together a system like this,and show that you are wrong-->he deals only in words and never in actual experimental setups as we do. So my challenge to him(and all here),show me a change in energy output or energy input in this system-on system 1,when system 2(the solar panel and resistive load)has a load placed on it.
Only words will follow,and very few experimental setups to see actual results.
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Hi Bill
I know what your saying,but think of it this way. The light bulb will emmit x amount of lumens of light regardless of wether the solar panel is there or not.The lumen output of the bulb will not change from the solar panel being open circuit,or has a load placed across it's output-->0 reflection shown from the solar panel in either case. The heat output is radiant,and that also dose not change regardless of the solar panel and its load. No matter how you look at it,the solar panel changes nothing in regards to the heat and light output of the bulb.The solar panel also will not show any reflection on the P/in to the light bulb.
That's not entirely true, but it is close enough not to quibble.
Lets calculate and messure this system as we would any other.
First P/in-->this being a DC system is easy-->volts x amps.This is our energy consumption.
Then calculate the energy output from the bulb. Heat and light. This is also easy,as it is equal to the energy in-->energy can be neither created nor destroyed-only transformed.We also must take into account the heat from the conecting wires and battery it self. As this will be a radiant heat,then it is an energy output.
So we have accounted for all our energy in and out,and have a net result of 0.
That is one isolated system and you have settled your balance.
Now system two-->our solar panel and resistive load(100 ohms in this case). This is also a DC output,so easy to calculate.
Note that the solar panel does not drive the resistor, and the resistor does not generate heat unless you supply energy to solar panel. So you have a couple of choices: Draw a boundary around the solar panel, or the solar panel and the resistor and EIN = EOUT + delta ESTORE for either of those systems. Or draw a boundary that includes the light bulb, and now the light that hits the solar panel from the light bulb does not leave the interior, being absorbed instead by the solar panel, and guess what? EIN = EOUT + delta ESTORE for the newly defined system.
So now,we do as we always do. We place the load on the solar panel,and then remeassure the energy output from system 1. Do you think it will change when the solar panel(system 2)has a load across it? If system 1 shows no change when system 2 is put into opperation,then system 2 is generating excess energy,as the energy in system 1 has been accounted for in the way of heat and light.
So in summery
System 2 dosnt change the heat output of system 1
System 2 dosnt change the light output of system 1
And system 2 dosnt reflect on the P/in of system 1
No, but it changes the available heat and light output of any system that includes System 1 and System 2. Surely you are not so naive as to believe that we could put another solar panel behind the first one and it would intercept the same light as the first.
If the solar panel was open(no load across it) then it would be said that it isnt consuming any energy,as energy cannot be destroyed.If nothing is comming out,then nothing is going in. So all you have to do now Bill,is place a load across that solar panel,and show me any change in either the energy in(watts) to system 1,or the energy out of system 1,and i'll believe what you say ;)
When solar panels don't have loads across them, the intinsic photo diodes convert the electricity to heat. Those photodiodes are what set the PV panel's open circuit voltage.
Whats happening here is a case of-->it's to simple to be true.
But if we had a transformer where the secondary shows no reflection on the primary what so ever,then every one would be jumping for joy. All we have here is that transformer,where light is converted into electrical energy,but shows no reflection on the primary.
First, there is a reflection, but it is so small that for all practical purposes we neglect it. Second, the more you decouple an energy source from its load, the less source energy reaches the load, IE if you count efficiency as output of the load divided by the energy of the input, the efficiency is very poor. That is for example what happens with Bill Alek's SFT transformer. In the case of the solar panel, the light converts a fraction of the input electricity to light in the wavelength range that the solar panel responds to. Then the solar panel covers only a small solid angle from the light. Then the solar panel has its own limited efficiency. By the time you are done only a tiny fraction of the energy supplied to the light comes out of the solar panel as electricity.
MarkE will throw all sorts of curve balls at you(eg.boundaries)but never put together a system like this,and show that you are wrong-->he deals only in words and never in actual experimental setups as we do. So my challenge to him(and all here),show me a change in energy output or energy input in this system-on system 1,when system 2(the solar panel and resistive load)has a load placed on it.
Only words will follow,and very few experimental setups to see actual results.
There is really no need to build such a silly experiment because it is blindingly obvious that there is no surplus energy to be had powering a light and directing that light at a PV module. Below is a diagram that shows just some possible boundaries that can be drawn around elements of this: battery, light, PV panel, resistor combination. No matter what combination of the four elements including all or none that one draws, when one accounts for all energy entering the interior through the boundary, and all energy exiting to the outside through the boundary: EIN - EOUT = delta ESTORED.
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Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.
A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.
Vidar
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Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.
A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.
Vidar
The end result of such a scheme is the equivalent of a heat engine operating with an efficiency dictated by the reservoirs at the highest and lowest temperatures. It does not and can not ever give rise to a self running system where the heat recovered by the heat pump is in excess of the heat required to drive the rest of the system.
The efficiency of the heat pump is at a maximum when the temperature of the hot and cold reservoir is at a minimum. The opposite is true for the heat engine. Combine those two facts together and the result given above can be deduced.
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Don't know if this is said before in this thread, but when we speak of open systems such as a heat pump, you put in x amount of work and get Y amount of heat out.
If the COP is 5, you get 5 times energy as heat out than the input work.
No you don't. For a COP 5 heat pump you get about 5/6ths useful heat energy moved as you put in. 1 part motive energy (usually electricity) moves 5 parts heat from a colder reservoir to a hotter reservoir. The 5 parts heat moved is the useful result. Absent the 5 parts heat in the colder reservoir the heat pump does not function. The cold reservoir only has to get cold enough that it no longer rejects heat to the refrigerant, or the hot reservoir so hot that the refrigerant cannot reject heat to it and the heat pump stops working.
A heat engine have the opposite efficiency. A heat engine is in principle a heat pump that is somewhat reversed. You put heat in, and get work out. This efficiency will allways be lower than 1/COP of the heat pump if you are using the only two reservoirs available on the heat pump - the cold and the hot one.
In order to make this heat engine to work as wanted, you need a third reservoir. Say the heat pump is using outside air as reservoir 1. Reservoir 2 is the heat pump output. Reservoir 3 is the ground.
Now, connect one side of the heat engine to the cold or hot reservoir of the heat pump, and connect the other reservoir to the ground. Then this heat engine will potentially do more work than the work required to run the heat pump. I discussed this issue with competent people at physicsforums.com. It seems reasonable.
Vidar
Similar unworkable ideas have been proposed many times. In order to make such a machine work, you have to extract more energy returning heat energy from the hot reservoir to a cold reservoir than you expend moving energy from a cold reservoir to the hot reservoir. That means that you not only have to convert all heat from a single reservoir to useful work, you have to more than do that. That puts you squarely at odds with the Second Law of Thermodynamics. So, first you will have to find an exception to the Second Law.
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There you go again with inffering things i never said
Surely you are not so naive as to believe that we could put another solar panel behind the first one and it would intercept the same light as the first.
But here is a little something for you to ponder. Take a large glass jug,and place one solar panel behind the other. In front of the first solar panel place a light bulb.Cover the glass jug so as no ambiant light can get into the jug/reflect onto the solar panels.Switch on the light bulb and messure the avaliable power from each solar panel. The second solar panel wont have any where near as much as the first solar panel(the one opposite the light bulb)Right?.
Now,fill jug with water,and redo test.
Your boundaries are junk. As can be seen in your diagram,you just place them where ever you think will help you debunk anyones idea's.
Please show me(and everyone here) that placeing a solar panel near a light bulb,and placing a load across that solar panel, reduces the output of the lightbulb,and/or increases the energy input to that light bulb.Show me this,and then we'll talk. You cannot. Insted you come up with all these mythical themes like boundaries.The !!boundary! is around each system-as i explained. When one system dose not impart or reflect on another,then that boundary is set.We have two sepperate system's,and one feeds off another without interference.
lets do a little more math here.
an incandecent bulb is about 5 to 6% efficient at converting electrical energy into light.
So a 5 watt bulb would put out around 300mW of light power.
A solar panel at best is about 17% efficient at converting light energy into electrical energy. So from our 300mW of light energy,we would see 17% of that converted into electrical energy.This means that we should only have around 51mW output from the solar panel at best if we cover the whole bulb in solar panel. So we need only achieve 1 volt over an 18 ohm load from our solar panel ;)
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So, first you will have to find an exception to the Second Law.
http://www.wsj.com/articles/SB1028220616493488320
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Tinman, using the concept of boundaries for analyzing these types of systems is absolutely real and is a common practice. Your problem is that when you encounter something new that you never heard of is that you will reject it out of hand. Then the concept has to be pushed and reinforced on you over and over until you eventually accept it.
Instead of just rejecting a concept because don't understand it and you want to reject it and you don't like it, why don't you accept it and then try to use the concept to advance your own argument? It's up to you to make an effort to learn instead of this push-back all the time. I have seen you reject legitimate concepts over and over just because you are not aware of them. Every time you put up a barrier, all that you do is slow yourself down.
It goes right back to five or more years ago. You were making very very simple circuits and demos and almost every time you were convinced that your latest demo was "something that conventional electrical engineering does not understand." It was never the case, ever. I think you know Mongrel Shark on YouTube. About 1 1/2 years ago I got into a discussion with him. He is a nice guy but he was just beginning his exploration in electronics. I said a few things to him that knocked him off balance because he wasn't understanding. He said something like "I don't have the time to teach you" and then he blocked me. He was making all of the classic "free energy electronics experimenter" mistakes because he was believing a lot of the BS myths out there. There is nothing more frustrating than seeing a guy that is so sure of himself but he may have only touched a scope for the first time three months before.
Before you discount the concept of setting up boundaries for analyzing systems, why don't you spend a few hours looking the material up online and reading about it and pulling yourself up by your own bootstraps? Your regular push-back only is to your own detriment. Mark is not perfect, but he really knows his stuff. Why should he have to expend 10 times the energy to get you to absorb a concept as compared to a conventional classroom setting where the teacher teaches the material once and then the student does his own follow-up work to learn and fully understand the material? The student may come back with some follow-up questions which is fine, but the student doesn't just sit there and push back.
MileHigh
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There you go again with inffering things i never said
The double counting is exactly what you are doing.But here is a little something for you to ponder. Take a large glass jug,and place one solar panel behind the other. In front of the first solar panel place a light bulb.Cover the glass jug so as no ambiant light can get into the jug/reflect onto the solar panels.Switch on the light bulb and messure the avaliable power from each solar panel. The second solar panel wont have any where near as much as the first solar panel(the one opposite the light bulb)Right?.
Now,fill jug with water,and redo test.
The point was that whatever incident light is intercepted by one solar panel is not available to another solar panel no matter what you do. If you construct optics to redirect light towards a second panel and/or redistribute light that gets to the first, whatever light that strikes the first panel after the reconfiguration will still not be available to the second panel, or any other load.
Your boundaries are junk. As can be seen in your diagram,you just place them where ever you think will help you debunk anyones idea's.
In order to evaluate energy balances, one needs to define a boundary around what they are talking about. Once they do, then the relationship: EIN - EOUT = delta ESTORED always holds. The diagram showed nine of the 16 possible boundaries. The other seven were left off as only a convenience to the drawing which is why the annotation states that the drawing only shows some of the combinations. You are free to establish any combination of elements and try to show an exception to the above relationship. When you correctly state that the battery and lamp are balanced, you neglect that balance requires that the energy lost from the battery must exit the combination of the battery and the lamp. That some of that energy makes its way to the solar panel and eventually the resistor does not create energy. The energy that goes to the solar panel from the lamp is not available to do anything else.Please show me(and everyone here) that placeing a solar panel near a light bulb,and placing a load across that solar panel, reduces the output of the lightbulb,and/or increases the energy input to that light bulb.Show me this,and then we'll talk. You cannot.
Why in the world do you think anyone needs to argue against your strawman? Insted you come up with all these mythical themes like boundaries.The !!boundary! is around each system-as i explained. When one system dose not impart or reflect on another,then that boundary is set.We have two sepperate system's,and one feeds off another without interference.
Are you of the strange opinion that anyone has been telling you that the earth's impact on sunlight is anything other than to: a) Cast a shadow in the forward direction of that sunlight, and b) scatter some of it off in other directions? All of this discussion came about from your assertion that gas in your apparatus is able to drive your ram without losing the energy that the ram imparts to the outside world. As has been explained to you in this discussion many times now, the gas loses the energy that does work in the outside system, just as surely as the sun loses the energy that it emits. That is when you offered your double counting of energy emitted by a battery powered lamp, and the electrical energy evolved from a solar panel exposed to that lamp.
lets do a little more math here.
an incandecent bulb is about 5 to 6% efficient at converting electrical energy into light.
So a 5 watt bulb would put out around 300mW of light power.
A solar panel at best is about 17% efficient at converting light energy into electrical energy. So from our 300mW of light energy,we would see 17% of that converted into electrical energy.This means that we should only have around 51mW output from the solar panel at best if we cover the whole bulb in solar panel. So we need only achieve 1 volt over an 18 ohm load from our solar panel ;)
What is this trivial exercise supposed to prove? Are you going to be amazed if you get say 60mW from a solar panel from 5W into your light bulb because you misinterpret the solar panel's specifications? You will still have a situation where every bit of heat and light emitted by the bulb was supplied by your electricity source. You will still have a situation where every bit of electricity and heat output from the solar panel is a result of the incident radiation striking it. You will still be no closer to creating support for your idea that you can perform external work with your ram without first expending that energy from the gas in the cylinder plus pressure vessel than when you started.
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Tinman, using the concept of boundaries for analyzing these types of systems is absolutely real and is a common practice. Your problem is that when you encounter something new that you never heard of is that you will reject it out of hand. Then the concept has to be pushed and reinforced on you over and over until you eventually accept it.
Instead of just rejecting a concept because don't understand it and you want to reject it and you don't like it, why don't you accept it and then try to use the concept to advance your own argument? It's up to you to make an effort to learn instead of this push-back all the time. I have seen you reject legitimate concepts over and over just because you are not aware of them. Every time you put up a barrier, all that you do is slow yourself down.
It goes right back to five or more years ago. You were making very very simple circuits and demos and almost every time you were convinced that your latest demo was "something that conventional electrical engineering does not understand." It was never the case, ever. I think you know Mongrel Shark on YouTube. About 1 1/2 years ago I got into a discussion with him. He is a nice guy but he was just beginning his exploration in electronics. I said a few things to him that knocked him off balance because he wasn't understanding. He said something like "I don't have the time to teach you" and then he blocked me. He was making all of the classic "free energy electronics experimenter" mistakes because he was believing a lot of the BS myths out there. There is nothing more frustrating than seeing a guy that is so sure of himself but he may have only touched a scope for the first time three months before.
Before you discount the concept of setting up boundaries for analyzing systems, why don't you spend a few hours looking the material up online and reading about it and pulling yourself up by your own bootstraps? Your regular push-back only is to your own detriment. Mark is not perfect, but he really knows his stuff. Why should he have to expend 10 times the energy to get you to absorb a concept as compared to a conventional classroom setting where the teacher teaches the material once and then the student does his own follow-up work to learn and fully understand the material? The student may come back with some follow-up questions which is fine, but the student doesn't just sit there and push back.
MileHigh
Thanks, but I think some pushback is critical and IMO should be encouraged. The road to hell is paved with expert opinion. I think it is misuse of authority that most turns people off about school. I know it pissed me off to no end.
Critical thinking demands that evidence rules the day. I have continued to engage Tinman because I think he really wants to learn the truth. I also harbor the hope that others watching his journey learn a thing or two from it: Better critical thinking skills, perhaps some better understanding of physics, and that learning is about gathering and testing data. It isn't about mindlessly accepting what anyone dictates no matter what their supposed authority might be. If a time comes that I conclude it is a lost cause, then I will simply stop engaging him.
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http://www.wsj.com/articles/SB1028220616493488320
That's step one. Next one has to find a way to employ that apparent exception. So far it's no dice.
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If you construct optics to redirect light towards a second panel and/or redistribute light that gets to the first, whatever light that strikes the first panel after the reconfiguration will still not be available to the second panel, or any other load. Once they do, then the relationship: EIN - EOUT = delta ESTORED always holds. The diagram showed nine of the 16 possible boundaries. The other seven were left off as only a convenience to the drawing which is why the annotation states that the drawing only shows some of the combinations. You are free to establish any combination of elements and try to show an exception to the above relationship. When you correctly state that the battery and lamp are balanced, you neglect that balance requires that the energy lost from the battery must exit the combination of the battery and the lamp. That some of that energy makes its way to the solar panel and eventually the resistor does not create energy. The energy that goes to the solar panel from the lamp is not available to do anything else.Why in the world do you think anyone needs to argue against your strawman?Are you of the strange opinion that anyone has been telling you that the earth's impact on sunlight is anything other than to: a) Cast a shadow in the forward direction of that sunlight, and b) scatter some of it off in other directions? All of this discussion came about from your assertion that gas in your apparatus is able to drive your ram without losing the energy that the ram imparts to the outside world. That is when you offered your double counting of energy emitted by a battery powered lamp, and the electrical energy evolved from a solar panel exposed to that lamp.
In order to evaluate energy balances, one needs to define a boundary around what they are talking about.
let me guess-it is you that will define this boundary? lol
You will still have a situation where every bit of heat and light emitted by the bulb was supplied by your electricity source.
That is correct,and that same accounting is true regardless of wether the solar panel is there or not. If i switch on a light in an empty room,the light will consume X amount of energy,and will disipate that same amount of energy. If i place a chair in the room,has any of the before mentioned values change in regard to energy in and out of that light bulb?-->no.
You will still have a situation where every bit of electricity and heat output from the solar panel is a result of the incident radiation striking it.
That is correct,but it still is an accounted for energy,and dose not reflect on the source.
You will still be no closer to creating support for your idea that you can perform external work with your ram without first expending that energy from the gas in the cylinder plus pressure vessel than when you started.
No energy is lost when the ram is used to perform work on the open system.The energy contained within the vessel and ram will remain the same regardless of wether a resistance is placed on that ram or not.
Are you going to be amazed if you get say 60mW from a solar panel from 5W into your light bulb because you misinterpret the solar panel's specifications?
Are you suggesting or implying that i didnt take into acount the wave lengths of light that the solar panel will opperate on,or the wave lengths of light that the incandecent bulb put's out? That would be your mistake.
As has been explained to you in this discussion many times now, the gas loses the energy that does work in the outside system, just as surely as the sun loses the energy that it emits.
There are no losses,only transformations-->maybe you should place some of your boundaries around these systems that you consider to be !!loosing!! energy,and then you will see that the energy is still there-->it's never lost.
The double counting is exactly what you are doing.The point was that whatever incident light is intercepted by one solar panel is not available to another solar panel no matter what you do.
Unless of course you have transparent solar panel's. Let me guess,you think there is no such thing?.
What is this trivial exercise supposed to prove?
The posabilities within.
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MileHigh
Tinman, using the concept of boundaries for analyzing these types of systems is absolutely real and is a common practice. Your problem is that when you encounter something new that you never heard of is that you will reject it out of hand. Then the concept has to be pushed and reinforced on you over and over until you eventually accept it.
Take a look at MarkE's diagram of these boundaries MH,and tell me which one of these boundaries are we to use on such a system?. Do we just pic and choose,or do we let the guru's pick the one that discounts any excess energy-->as this happens far to often. The boundary between two systems is set where as one system dose not impart or influence the other in any way.
Instead of just rejecting a concept because don't understand it and you want to reject it and you don't like it, why don't you accept it and then try to use the concept to advance your own argument?
As above-->which boundary shall i use out of the many presented.
It's up to you to make an effort to learn instead of this push-back all the time. I have seen you reject legitimate concepts over and over just because you are not aware of them. Every time you put up a barrier, all that you do is slow yourself down.
Your legitimate concepts are just word's. No one has shown me yet these laws being applied to the systems i present. What has been adopted here is the !one fits all! bullshit.
It goes right back to five or more years ago. You were making very very simple circuits and demos and almost every time you were convinced that your latest demo was "something that conventional electrical engineering does not understand." It was never the case, ever. I think you know Mongrel Shark on YouTube. About 1 1/2 years ago I got into a discussion with him. He is a nice guy but he was just beginning his exploration in electronics. I said a few things to him that knocked him off balance because he wasn't understanding. He said something like "I don't have the time to teach you" and then he blocked me. He was making all of the classic "free energy electronics experimenter" mistakes because he was believing a lot of the BS myths out there. There is nothing more frustrating than seeing a guy that is so sure of himself but he may have only touched a scope for the first time three months before.
We all have our strong points and weak point's. I am an engineer by trade,and have worked with pneumatic systems for many year's,and i have seen gas temperatures rise within pneumatic cyliders to be far greater than the temperature of the supply gas when that cylider is made to do work. If there is a rise in temperature/pressure and volume,then there is a rise in energy-->not a fall.All the test i have seen in regards to these ideal gas laws are not even close to being performed in a way that represents this situation. This is another !one fits all! claim,and it simply isnt true.
Before you discount the concept of setting up boundaries for analyzing systems, why don't you spend a few hours looking the material up online and reading about it and pulling yourself up by your own bootstraps? Your regular push-back only is to your own detriment. Mark is not perfect, but he really knows his stuff. Why should he have to expend 10 times the energy to get you to absorb a concept as compared to a conventional classroom setting where the teacher teaches the material once and then the student does his own follow-up work to learn and fully understand the material? The student may come back with some follow-up questions which is fine, but the student doesn't just sit there and push back.
We were taught many things in school that do not reflect actual events. Take for instance the earth orbiting around the sun-->how wrong was that. How in hell can the earth orbit around a sun that is moving. So you can listen to a teacher that teaches incorrect description's,or you listen and learn on the job. When some one is trying to tell you that what you have been seeing right before your eyes for the past 13 years is incorrect,then it's time for a new teacher.
From guru's(most) we see word salads,googlegasms and palava-->but we never see any actual devices being tested by said people. Poynt and TK are what i would call !the teachers that make an effort!,the rest just rely on what they read,that really has nothing to do with the DUT-->its just that!one fits all! rubbish. Well one dosnt fit all,and a ram doing work is one of those situations where one dosnt fit all. When you are going to use some scientific method to determond a systems outcome,then make sure that data was obtained around the DUT-->not some other device that dosnt represent your DUT.
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let me guess-it is you that will define this boundary? lol
If you think that you have not been paying attention. I did not say "I". I did not say some authority. I said "one". For your device, it is you who needs to define what is inside whatever collection of things that together you wish to evaluate and what is the outside world.
That is correct,and that same accounting is true regardless of wether the solar panel is there or not. If i switch on a light in an empty room,the light will consume X amount of energy,and will disipate that same amount of energy. If i place a chair in the room,has any of the before mentioned values change in regard to energy in and out of that light bulb?-->no.
For all practical purposes, of course.
That is correct,but it still is an accounted for energy,and dose not reflect on the source.
This idea of thinking that there is a strong reflection is a strawman of your invention. No one has argued such a position.
No energy is lost when the ram is used to perform work on the open system.The energy contained within the vessel and ram will remain the same regardless of wether a resistance is placed on that ram or not.
The system consisting of the pressure vessel and the ram cylinder absolutely gives up the work transferred to the outside world. This is the key point that you have been missing. It happens just as the battery transfers energy to the lamp, and the lamp to space surrounding it, and the solar panel to the resistor that loads it. I accept that you will continue to reject this despite countless references backing it up until you perform your experiments with proper measurements.
Are you going to be amazed if you get say 60mW from a solar panel from 5W into your light bulb because you misinterpret the solar panel's specifications?
Are you suggesting or implying that i didnt take into acount the wave lengths of light that the solar panel will opperate on,or the wave lengths of light that the incandecent bulb put's out? That would be your mistake.
Since you only stated a set of presumed coefficients and a presumed result as part of an argument I could only guess as to where you intend to go.
There are no losses,only transformations-->maybe you should place some of your boundaries around these systems that you consider to be !!loosing!! energy,and then you will see that the energy is still there-->it's never lost.
Are you having trouble with the word "lost"? I did not say "destroyed". Electricity converted to heat is a loss. The energy is still there, you can't make or destroy the stuff, but it is no longer in a useful form.
Unless of course you have transparent solar panel's. Let me guess,you think there is no such thing?.
The posabilities within.
You do understand don't you that every bit of light energy and then some that any one such panel converts to electricity does not continue on as light?
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The end result of such a scheme is the equivalent of a heat engine operating with an efficiency dictated by the reservoirs at the highest and lowest temperatures. It does not and can not ever give rise to a self running system where the heat recovered by the heat pump is in excess of the heat required to drive the rest of the system.
Agree. That is what I meant.
The efficiency of the heat pump is at a maximum when the temperature of the hot and cold reservoir is at a minimum. The opposite is true for the heat engine. Combine those two facts together and the result given above can be deduced.
Agree, and thanks for the simple, but well put explanation :-)
Vidar
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. I did not say "destroyed". but it is no longer in a useful form.
If you think that you have not been paying attention. I did not say "I". I did not say some authority. I said "one". For your device, it is you who needs to define what is inside whatever collection of things that together you wish to evaluate and what is the outside world.
And this boundary i have set !how many! times now?.
For all practical purposes, of course.This idea of thinking that there is a strong reflection is a strawman of your invention.
And again-->what reflection. Please copy and paste this reflection i spoke off. It was you that brought up the reflection issue,not i.
.The system consisting of the pressure vessel and the ram cylinder absolutely gives up the work transferred to the outside world.
I think you have the wrong picture set in your head about this system,so i will try once again to explain.
The vessel supplies the compressed gas to the ram. First test is performed with no resistance placed apon the ram. The valve is opened,and the ram extends to it's full travel without resistance(other than the frictional resistance of the piston seals against the cylinder wall). The pressure avaliable to the ram is set at say 50psi gauge pressure via a regulator(<-- important to remember).
So the pressure in the ram only reaches it's maximum value once the ram has traveled it's full distance. We can now take a temperature reading of the gas in the vessel and ram,along with a pressure reading of the vessel-->we already know the ram will have a pressure of 50psi,as it is regulated.
Test 2-Now we run the test again,but this time a high resistance is placed apon the ram.So now the ram reaches it's maximum pressure at the begining of it's travel-not the end.Once the ram has traveled to it's limit,you once again take a temperature reading of the gas inside the ram and vessel,and also the pressure of the gas inside the vessel-->we know the ram pressure is 50psi.
At this point,all your ideal gas laws and boyles law fall apart,because you now find that the temperature of the gas inside the ram cylinder is higher in test 2 than it was in test 1.You also find that the pressure in the supply vessel hasnt dropped as much in test 2 as it did in test 1. The reason for this is that the gas temperature in the ram in test 2 is higher,and thus the gas has expanded more so than it did in test 1,and so not as much gas was required from the vessel in test 2 as it was in test one. So now your scratching your head,trying to work out how making that gas do work actually used less avaliable energy within the gas than that as if there were no work done by the gas at all. Now i know (without doubt)that you will argue with this,as it dosnt comply with your wonderful law's,and the reason for this is because the !one fit's all! just dosnt fit all.
As i stated before,i have seen this happen for over 13 years on the job. Maybe you should ring the engineer and maintenance department at BHP,and tell them all there temperature sensors are reading wrong. How is it that the supply gas temperature can be 76*C(from the vessel),and yet the gas temperature within the ram's can reach a temperature of over 100*C-->at which point the coolers come on.You see the same thing in electrical flow-->what happens when you add a resistance-->yes,you generate heat.
Are you having trouble with the word "lost"?
Not at all. I think it more a case that you are having trouble with correct outcomes of different system's,as you insist that the !one fits all! stand's-->it dose not. It's much the same as the planets orbit the sun-->they do not.
Since you only stated a set of presumed coefficients and a presumed result as part of an argument I could only guess as to where you intend to go.
And yet you were happy to insinuate that-Quote: because you misinterpret the solar panel's specifications?
So you say that i misinterpret,and yet you were going on presumptions.
Electricity converted to heat is a loss.
Not if it's heat you want-->and if you go back to the beginig of this thread,you will see that heat is one of the very things we are trying to generate-->as is the case with the ram experiment.
You do understand don't you that every bit of light energy and then some that any one such panel converts to electricity does not continue on as light?
I do. But are you able to calculate the energy from that small portion of light that is !lost!when the solar panel is within that light? If i replace the solar panel with a piece of wood the same size,we still loose that small amount of light. How much energy can we generate from that small piece of wood?.
The energy is still there, you can't make or destroy the stuff,
You know this for sure?. I could ask as to where the energy within the universe came from in the begining,but i think that is beyond any ones means to explain. Did it just appear out of nothing?
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And this boundary i have set !how many! times now?.
You have counted energy crossing in and crossing out using the same sign: IE double counting.
And again-->what reflection. Please copy and paste this reflection i spoke off. It was you that brought up the reflection issue,not i.
Really? Where and when did I make such a claim?
.I think you have the wrong picture set in your head about this system,so i will try once again to explain.
The vessel supplies the compressed gas to the ram. First test is performed with no resistance placed apon the ram. The valve is opened,and the ram extends to it's full travel without resistance(other than the frictional resistance of the piston seals against the cylinder wall). The pressure avaliable to the ram is set at say 50psi gauge pressure via a regulator(<-- important to remember).
So the pressure in the ram only reaches it's maximum value once the ram has traveled it's full distance. We can now take a temperature reading of the gas in the vessel and ram,along with a pressure reading of the vessel-->we already know the ram will have a pressure of 50psi,as it is regulated.
Problem #1. You need to know the pressure, temperature, volume and gas composition of both volumes before you open the valve. That tells you the starting energy.
Problem #2. Regulators only regulate if there is a gas flow. A restriction only creates a pressure drop when there is flow. So, the regulator does not guarantee pressure. You need pressure gauges / sensors to know the pressure.
Problem #3. You need to know the pressure, temperature, volume and gas composition of both volumes immediately after you finish the stroke. That tells you the ending energy.
Test 2-Now we run the test again,but this time a high resistance is placed apon the ram.So now the ram reaches it's maximum pressure at the begining of it's travel-not the end.Once the ram has traveled to it's limit,you once again take a temperature reading of the gas inside the ram and vessel,and also the pressure of the gas inside the vessel-->we know the ram pressure is 50psi.
At this point,all your ideal gas laws and boyles law fall apart,because you now find that the temperature of the gas inside the ram cylinder is higher in test 2 than it was in test 1.You also find that the pressure in the supply vessel hasnt dropped as much in test 2 as it did in test 1.
GIGO. You didn't know the energy when you started. You didn't determine the energy when you ended because you incorrectly assumed that the pressure was the regulator setting. So go back to Square One, instrument so that you can determine the actual starting and ending energies and then see if you still think you observe work being done for free.
The reason for this is that the gas temperature in the ram in test 2 is higher,and thus the gas has expanded more so than it did in test 1,and so not as much gas was required from the vessel in test 2 as it was in test one. So now your scratching your head,trying to work out how making that gas do work actually used less avaliable energy within the gas than that as if there were no work done by the gas at all. Now i know (without doubt)that you will argue with this,as it dosnt comply with your wonderful law's,and the reason for this is because the !one fit's all! just dosnt fit all.
Like I said: From a measurement standpoint it is GIGO.
As i stated before,i have seen this happen for over 13 years on the job. Maybe you should ring the engineer and maintenance department at BHP,and tell them all there temperature sensors are reading wrong. How is it that the supply gas temperature can be 76*C(from the vessel),and yet the gas temperature within the ram's can reach a temperature of over 100*C-->at which point the coolers come on.
The reason is well known: You are removing energy from the supply and putting it into the ram gas volume.You see the same thing in electrical flow-->what happens when you add a resistance-->yes,you generate heat.
Is your thesis here that more resistance or less resistance gives rise to the most heat evolution in the resistor? Any amount of resistance less than infinity but greater than zero evolves more heat in the resistor than an infinite or zero resistance. Any value less than infinity removes energy from the source.Not at all. I think it more a case that you are having trouble with correct outcomes of different system's,as you insist that the !one fits all! stand's-->it dose not. It's much the same as the planets orbit the sun-->they do not.
I am sorry, but you need to review your Kepler.And yet you were happy to insinuate that-Quote: because you misinterpret the solar panel's specifications?
If you think that you measure an anomaly in the electricity output that would be due to a misinterpretation of the specifications. I stand by my statement.So you say that i misinterpret,and yet you were going on presumptions.
Actually it's your experiment's presumptions that have lead you to GIGO.Not if it's heat you want-->and if you go back to the beginig of this thread,you will see that heat is one of the very things we are trying to generate-->as is the case with the ram experiment.
I do. But are you able to calculate the energy from that small portion of light that is !lost!when the solar panel is within that light? If i replace the solar panel with a piece of wood the same size,we still loose that small amount of light. How much energy can we generate from that small piece of wood?.
Solid geometry is well understood.You know this for sure?. I could ask as to where the energy within the universe came from in the begining,but i think that is beyond any ones means to explain. Did it just appear out of nothing?
Once again you are not paying attention. I have explained the difference between "loss" and "destroyed". Setting up strawman arguments does not help your case.
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Where and when did I make such a claim?
Post 257-Quote: First, there is a reflection, but it is so small that for all practical purposes we neglect it.
So, the regulator does not guarantee pressure. You need pressure gauges / sensors to know the pressure.
You must have a crappy regulator. Mine has a pressure gauge on both the supply and delivery side.Once the delivery side has reached the set value,gas discontinues to flow.The pressure holds stedy at the preset value.The pressure gauge on the delivery side of the regulator shows actual cylinder pressure. So the cylinder/ram dose have a pressure gauge,and the regulator will hold stedy at the set value of 50psi regardless of flow.
You need to know the pressure, temperature, volume and gas composition of both volumes before you open the valve.
This is just common sense,and it is why i never included this in the quick test sketchup.When calculating energy gains or losses,of course we measure what we have in the begining.
You need to know the pressure, temperature, volume and gas composition of both volumes immediately after you finish the stroke.
As i have stated before.__> this we know.
Regulators only regulate if there is a gas flow
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Incorrect. Regulators regulate maximum avaliable pressure.Once that set pressure has been reached within the ram,the gas will discontinue to flow.
A restriction only creates a pressure drop when there is flow.
We are not restricting flow,we are regualting pressure
That tells you the ending energy. GIGO. You didn't know the energy when you started
As stated above-this is common sense. Of course we calculate the avaliable energy at the start of the test. Spoon feeding is starting to come to mind here :o
You didn't determine the energy when you ended because you incorrectly assumed that the pressure was the regulator setting
You need to get your self a decent regulator that actually regulates set maximum pressure.
So go back to Square One, instrument so that you can determine the actual starting and ending energies and then see if you still think you observe work being done for free.
Already covered above. I really dont think spoon feeding is required ATM.
You are removing energy from the supply and putting it into the ram gas volume.
Indeed we are. And as the system is still closed,the energy loss is where? We have a 10 ltr bucket that has 5 ltr's of water in it.we tip 2 ltr's into another bucket-->how much water do we now have in the two buckets in total?.
Any value less than infinity removes energy from the source.
No,it dose not !remove! any energy,it transforms energy.
I am sorry, but you need to review your Kepler.
No thanks,i will stick with actual events and outcomes. Are you really saying that i will loose more gas energy in test 2,than i would in test 1 :o. Lol,think i'll leave the books with you on this one,and stick to what ihave observed over the last 13 years-->actual facts.
Setting up strawman arguments does not help your case.
And yet here you are,useing the !one fits all! approach. So i'll ask again-->please show the experimental data recorded around the system i present. Also,as i have asked on a couple of occasions now-->are you able to calculate joules of energy avaliable in a gas volume of set pressure and temperature-->the gas being ambiant air.
Example.Our vessel has a 20ltr volume. Our gas pressure will be set at 100psi within that vessel,and the gas temperature will be 28*C.
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So now you don't understand the word "negligible"???
Anyway so I take it we can look forward to your carefully obtained data that shows you can perform external work without losing that energy from the source gas volume. Is that correct?
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So now you don't understand the word "negligible"???
I do. Do you accept the fact they you did indeed say there is a reflection.
Anyway so I take it we can look forward to your carefully obtained data that shows you can perform external work without losing that energy from the source gas volume. Is that correct?
You just havnt been reading things through-have you.You have this vision stuck in your head,and regardless of what i say,or how many times i explain the setup,your vision will not change.
Once again.
Regardless of wether the ram has a resistance against it(doing useful work)or not,the energy within the gas dose not drop,it in fact increases. Please re read post 271 reply 3.
Quote:The vessel supplies the compressed gas to the ram. First test is performed with no resistance placed apon the ram. The valve is opened,and the ram extends to it's full travel without resistance(other than the frictional resistance of the piston seals against the cylinder wall). The pressure avaliable to the ram is set at say 50psi gauge pressure via a regulator(<-- important to remember).
So the pressure in the ram only reaches it's maximum value once the ram has traveled it's full distance. We can now take a temperature reading of the gas in the vessel and ram,along with a pressure reading of the vessel-->we already know the ram will have a pressure of 50psi,as it is regulated.
Test 2-Now we run the test again,but this time a high resistance is placed apon the ram.So now the ram reaches it's maximum pressure at the begining of it's travel-not the end.Once the ram has traveled to it's limit,you once again take a temperature reading of the gas inside the ram and vessel,and also the pressure of the gas inside the vessel-->we know the ram pressure is 50psi.
I dont think the 2 test perameters could be set out much more clearly.
And again i ask-Quote post 273-
Also,as i have asked on a couple of occasions now-->are you able to calculate joules of energy avaliable in a gas volume of set pressure and temperature-->the gas being ambiant air.
Example.Our vessel has a 20ltr volume. Our gas pressure will be set at 100psi within that vessel,and the gas temperature will be 28*C.
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The whole idea her is to show that secondary systems(open systems) can feed of the energy from the primary system with out there being any effect on the primary system-yes,no equal and opposite reaction.This is where those boundaries come into it,and they are set between the device that dose not reflect on the other,and the other device. One is coupled to the other,but the other is not coupled to the first-hence the solar panel and light experiment.
Have you ever put a light bulb inside a reflective ball(like a ball that has a mirror surface all around it's surface on the inside)and tried to cast a shadow inside that ball?.
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The whole idea her is to show that secondary systems(open systems) can feed of the energy from the primary system with out there being any effect on the primary system-yes,no equal and opposite reaction.
Do you understand that equal and opposite reaction: Newton's Third Law is about force and not energy? The equal and opposite reactions occur where material/energy leaves one thing, and then again where it impacts another.
This is where those boundaries come into it,and they are set between the device that dose not reflect on the other,and the other device. One is coupled to the other,but the other is not coupled to the first-hence the solar panel and light experiment.
Have you ever put a light bulb inside a reflective ball(like a ball that has a mirror surface all around it's surface on the inside)and tried to cast a shadow inside that ball?.
No, I can't say that I ever attempted to amuse myself by trying to do that.
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I do. Do you accept the fact they you did indeed say there is a reflection.
If you understand "negligible" then you understand that means that I described the reflection as insignificant. Nothing except empty space is 100.0% emissive. Everything reflects at least a little. But for ever situation we have discussed the reflection is negligible. IE it does not affect the matters under discussion. IE you have been creating an artificial argument with yourself.
You just havnt been reading things through-have you.You have this vision stuck in your head,and regardless of what i say,or how many times i explain the setup,your vision will not change.
Once again.
Regardless of wether the ram has a resistance against it(doing useful work)or not,the energy within the gas dose not drop,it in fact increases. Please re read post 271 reply 3.
So let's be painfully clear here about what it is you claim:
1) You have some amount of energy in your combined gas volume when the ram is at an initial position X1: Let's label this so that we can keep track: UGAS_X1
2) The ram moves from position X1 to position X2 against an opposing force: FX1 to FX2. The X and F values are positive: The movement performs external work rather than absorbs work from the outside. Let's label the work done: WEXTERNAL.
3) The energy in the gas after the stroke has been executed is: UGAS_X2.
4) You contend that:
a) UGAS_X2 > UGAS_X1, AND
b) WEXTERNAL > 0, AND therefore:
c) UGAS_X2 + WEXTERNAL > UGAS_X1
Quote:The vessel supplies the compressed gas to the ram. First test is performed with no resistance placed apon the ram. The valve is opened,and the ram extends to it's full travel without resistance(other than the frictional resistance of the piston seals against the cylinder wall). The pressure avaliable to the ram is set at say 50psi gauge pressure via a regulator(<-- important to remember).
So the pressure in the ram only reaches it's maximum value once the ram has traveled it's full distance. We can now take a temperature reading of the gas in the vessel and ram,along with a pressure reading of the vessel-->we already know the ram will have a pressure of 50psi,as it is regulated.
Where are the temperature sensors located? What is their time constant? Are you taking a single reading or logging?
Test 2-Now we run the test again,but this time a high resistance is placed apon the ram.So now the ram reaches it's maximum pressure at the begining of it's travel-not the end.Once the ram has traveled to it's limit,you once again take a temperature reading of the gas inside the ram and vessel,and also the pressure of the gas inside the vessel-->we know the ram pressure is 50psi.
Same questions as for Test 1.
I dont think the 2 test perameters could be set out much more clearly.
And again i ask-Quote post 273-
Also,as i have asked on a couple of occasions now-->are you able to calculate joules of energy avaliable in a gas volume of set pressure and temperature-->the gas being ambiant air.
Example.Our vessel has a 20ltr volume. Our gas pressure will be set at 100psi within that vessel,and the gas temperature will be 28*C.
I have already told you that there is no great difficulty calculating energy in a dry gas volume of known composition, temperature, pressure, and volume.
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I do. Do you accept the fact they you did indeed say there is a reflection.
You just havnt been reading things through-have you.You have this vision stuck in your head,and regardless of what i say,or how many times i explain the setup,your vision will not change.
Once again.
Regardless of wether the ram has a resistance against it(doing useful work)or not,the energy within the gas dose not drop,it in fact increases. Please re read post 271 reply 3.
Quote:The vessel supplies the compressed gas to the ram. First test is performed with no resistance placed apon the ram. The valve is opened,and the ram extends to it's full travel without resistance(other than the frictional resistance of the piston seals against the cylinder wall). The pressure avaliable to the ram is set at say 50psi gauge pressure via a regulator(<-- important to remember).
So the pressure in the ram only reaches it's maximum value once the ram has traveled it's full distance. We can now take a temperature reading of the gas in the vessel and ram,along with a pressure reading of the vessel-->we already know the ram will have a pressure of 50psi,as it is regulated.
Test 2-Now we run the test again,but this time a high resistance is placed apon the ram.So now the ram reaches it's maximum pressure at the begining of it's travel-not the end.Once the ram has traveled to it's limit,you once again take a temperature reading of the gas inside the ram and vessel,and also the pressure of the gas inside the vessel-->we know the ram pressure is 50psi.
I dont think the 2 test perameters could be set out much more clearly.
And again i ask-Quote post 273-
Also,as i have asked on a couple of occasions now-->are you able to calculate joules of energy avaliable in a gas volume of set pressure and temperature-->the gas being ambiant air.
Example.Our vessel has a 20ltr volume. Our gas pressure will be set at 100psi within that vessel,and the gas temperature will be 28*C.
The thermodynamics of pressurising and stroking a pneumatic ram from a "constant pressure" supply is perhaps a little more complex than you realise .
However by placing appropriate boundaries and analysing over those those we can deduce why the exhaust air is at greater than ambient and why ram heats up, but does not break the known fact that air cools and looses heat while it expands.
Consider the system of the entire air supply AND ram as a closed system (not 'isolated') and that system does work on via the ram on an external system. We know that as the ram cylinder expands the volume of this closed system expands and thus the pressure must drop. Recall the amount of matter in the whole 'supply + ram system is fixed as per defining it as a closed system.
Since you specify the supply to the ram is 'constant pressure' we must either have a supply at a higher pressure that drops as air flows through a constant pressure valve OR we do work on the supply to maintain the pressure as the air expands into the ram cylinder.
This requirement means that to maintain a 'constant pressure' supply (as you specify) we must ADD energy into the supply as
we fill the ram. Where does this energy go to? It turns up as increasing temperature of the air in the volume we are filling in the ram, since compressing air causes the temperature to rise.
In a real system this energy would need to be added by a compressor pressurising the supply tank.
As the ram is allowed to expand against a load this air at elevated temperature cools. However Carnot sets the maximum efficiency on the conversion of heat to work and thus the exhaust air will be higher than ambient.
You are misinterpreting the apparent rise in temperature as some property of the expanding air, but as you see above the opposite is true.
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The thermodynamics of pressurising and stroking a pneumatic ram from a "constant pressure" supply is perhaps a little more complex than you realise .
However by placing appropriate boundaries and analysing over those those we can deduce why the exhaust air is at greater than ambient and why ram heats up, but does not break the known fact that air cools and looses heat while it expands.
Consider the system of the entire air supply AND ram as a closed system (not 'isolated') and that system does work on via the ram on an external system. We know that as the ram cylinder expands the volume of this closed system expands and thus the pressure must drop. Recall the amount of matter in the whole 'supply + ram system is fixed as per defining it as a closed system.
Since you specify the supply to the ram is 'constant pressure' we must either have a supply at a higher pressure that drops as air flows through a constant pressure valve OR we do work on the supply to maintain the pressure as the air expands into the ram cylinder.
This requirement means that to maintain a 'constant pressure' supply (as you specify) we must ADD energy into the supply as
we fill the ram. Where does this energy go to? It turns up as increasing temperature of the air in the volume we are filling in the ram, since compressing air causes the temperature to rise.
In a real system this energy would need to be added by a compressor pressurising the supply tank.
As the ram is allowed to expand against a load this air at elevated temperature cools. However Carnot sets the maximum efficiency on the conversion of heat to work and thus the exhaust air will be higher than ambient.
You are misinterpreting the apparent rise in temperature as some property of the expanding air, but as you see above the opposite is true.
LE
There is no further air supplied to the supply vessel during the test. The supply vessel has a high enough pressure and large enough capacity to supply the ram with the volume required to complete the operation of the ram.
As the system is closed, if the temperature is higher in the ram after test 2 than it was after test one, then which gas volume as a whole would be said to have more energy?
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LE
There is no further air supplied to the supply vessel during the test. The supply vessel has a high enough pressure and large enough capacity to supply the ram with the volume required to complete the operation of the ram.
As the system is closed, if the temperature is higher in the ram after test 2 than it was after test one, then which gas volume as a whole would be said to have more energy?
You completely miss the point. Even if the supply vessel is 'enormous' the pressure in it will drop as it expands into the cylinder.
For the combined supply/ram closed system we have defined to remain energy neutral you MUST do work on the system if the
supply to the ram is maintained at constant pressure. (other wise you'd be breaking the first law of thermodynamics and I don't believe you are claiming a first law violation).
In your case the supply tank is large, the pressure in it drops and it is replenished later with a compressor. The fact the compressor runs afterwards is not important from an energy standpoint. The amount of work to bring the supply back to the starting pressure is the equal (or more than) the amount of energy in the charge of air added to the ram.
The result is simple. The ram heats up as the air is compressed into it. As it expands it cools, but it does not cool to ambient due to the limitations of the Carnot efficiency.
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What it is looking like more and more is that you have a problem with your measurements. My guess is that it is a problem with the measurement resolution particularly of the pressure. If it takes n1 molecules from the pressure vessel to extend the ram under some circumstance and there were originally nPV_START molecules in the tank, then without any energy loss, the new pressure would be:
PPV_END = PPV_START*nPV_START/(n1+nPV_START), which is to say a difference of:
PPV_START - PPV_END = PPV_START * n1/(n1+nPV_START).
Just for a rough estimate: If PPV_START = 120psi, the pressure vessel is 2 cu ft, while the ram volume is say 0.005 cu ft retracted and 0.05 cu ft extended then:
You need 0.045 cu ft @ 50psi, and 0.005 cu ft @ (50-15)psi, and you will be getting that from 2 cu ft at 120 psi:
n1 = KGAS*2.425 cu ft psi
nPV_START = KGAS*240 cu ft psi
PPV_END = 120psi * 240/(242.425) = 118.8psi ETA corrected from nPV_END
PPV_START - PPV_END ~= 1.2psi ETA corrected from nPV_START - nPV_END
If you cannot resolve to about 1psi you won't see any change at all: Woo hoo free energy! If you cannot resolve to a fraction of a psi then you will get an optimistic pressure reading and again appear to have more energy at the end than when you started, all for free.
If you are looking for a pressure difference in a vessel where the temperature has changed, say by energy added by heating, and the temperature changes by 10C, then at room temperature the pressure only increases by (308K - 298K)/298K ~3%. In order to detect the pressure change and calculate the energy change within +/-10% you would need to resolve pressure to better than +/-0.34 psi accuracy.
The bigger that the pressure vessel is compared to the ram extended volume, and the higher the pressure in the pressure vessel compared to that in the ram, the greater the precision required in the pressure measurements in order to get any kind of sane result.
ETA: You can look at it as a problem of measuring energy removed from a water reservoir by weighing the reservoir. Unless you have super resolution instruments you just won't see the weight difference when withdrawing small fractions of the water volume. So, if you rely on those weight measurements: it may well seem as though you can have your water in the reservoir, and consume it at the same time. Compare that to what you believe your Test 1 and Test 2 data is telling you that you can do with energy from the pressure vessel.
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You completely miss the point. Even if the supply vessel is 'enormous' the pressure in it will drop as it expands into the cylinder.
This we know.
For the combined supply/ram closed system we have defined to remain energy neutral you MUST do work on the system if the
supply to the ram is maintained at constant pressure. (other wise you'd be breaking the first law of thermodynamics and I don't believe you are claiming a first law violation).
Im not quite sure what you mean here LE,but if i read correct,this is my reply.
The supply vessel's capacity and pressure is much higher than that of the ram. So a constant pressure lower than that of the supply vessel,and a volume much less than the supply vessel,we are able to maintain a constant pressure to the ram without doing work on the system/supply vessel. It could also be said that when the gas is doing work against a resistance,that resistance is also doing work against the gas.
In your case the supply tank is large, the pressure in it drops and it is replenished later with a compressor. The fact the compressor runs afterwards is not important from an energy standpoint. The amount of work to bring the supply back to the starting pressure is the equal (or more than) the amount of energy in the charge of air added to the ram.
With this i agree. The compressor to the supply tank is nothing more than a battery charger is to a battery. An air compressor is one of the most inefficient devices around,and you would be lucky to turn 20% of the energy used by the compressor into gas energy.
The result is simple. The ram heats up as the air is compressed into it. As it expands it cools, but it does not cool to ambient due to the limitations of the Carnot efficiency.
Do you mean as the ram expands the gas inside cool's?. If so,this is not the case,as the pressure is always increasing until it reaches it's regulated pressure.(see original diagram at start of thread)The reason the pressure continually rises is because the return springs are progresive. This remains the case regardless of wether there is a heavier resistance placed on the ram or not.
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What it is looking like more and more is that you have a problem with your measurements. My guess is that it is a problem with the measurement resolution particularly of the pressure. If it takes n1 molecules from the pressure vessel to extend the ram under some circumstance and there were originally nPV_START molecules in the tank, then without any energy loss, the new pressure would be:
PPV_END = PPV_START*nPV_START/(n1+nPV_START), which is to say a difference of:
PPV_START - PPV_END = PPV_START * n1/(n1+nPV_START).
Just for a rough estimate: If PPV_START = 120psi, the pressure vessel is 2 cu ft, while the ram volume is say 0.005 cu ft retracted and 0.05 cu ft extended then:
You need 0.045 cu ft @ 50psi, and 0.005 cu ft @ (50-15)psi, and you will be getting that from 2 cu ft at 120 psi:
n1 = KGAS*2.425 cu ft psi
nPV_START = KGAS*240 cu ft psi
PPV_END = 120psi * 240/(242.425) = 118.8psi ETA corrected from nPV_END
PPV_START - PPV_END ~= 1.2psi ETA corrected from nPV_START - nPV_END
If you cannot resolve to about 1psi you won't see any change at all: Woo hoo free energy! If you cannot resolve to a fraction of a psi then you will get an optimistic pressure reading and again appear to have more energy at the end than when you started, all for free.
If you are looking for a pressure difference in a vessel where the temperature has changed, say by energy added by heating, and the temperature changes by 10C, then at room temperature the pressure only increases by (308K - 298K)/298K ~3%. In order to detect the pressure change and calculate the energy change within +/-10% you would need to resolve pressure to better than +/-0.34 psi accuracy.
The bigger that the pressure vessel is compared to the ram extended volume, and the higher the pressure in the pressure vessel compared to that in the ram, the greater the precision required in the pressure measurements in order to get any kind of sane result.
ETA: You can look at it as a problem of measuring energy removed from a water reservoir by weighing the reservoir. Unless you have super resolution instruments you just won't see the weight difference when withdrawing small fractions of the water volume. So, if you rely on those weight measurements: it may well seem as though you can have your water in the reservoir, and consume it at the same time. Compare that to what you believe your Test 1 and Test 2 data is telling you that you can do with energy from the pressure vessel.
The resolution of the digital gauges i am useing is .1psi for pressure,and .1*C for temperature.
There is another way to test such a device useing only a starting pressure and temperature-->and these are only needed to insure each test starts with the same perameters/stored energy..
Can you guess what this test may be Mark-LE ? This test is indisputable.
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This we know.
Im not quite sure what you mean here LE,but if i read correct,this is my reply.
He means that the product of pressure and volume equals potential energy. If you increase the total volume by moving the ram at constant pressure then you are increasing the energy stored in the ram cylinder. If simultaneously, the pressure and volume of the pressure vessel remain fixed, then the energy in the pressure vessel is not changing and you must supply the energy that you are adding to the ram cylinder from somewhere else.
The supply vessel's capacity and pressure is much higher than that of the ram. So a constant pressure lower than that of the supply vessel,and a volume much less than the supply vessel,we are able to maintain a constant pressure to the ram without doing work on the system/supply vessel. It could also be said that when the gas is doing work against a resistance,that resistance is also doing work against the gas.
Just taking the gas statics and ignoring thermodynamics, that does not work. If the energy is constant, then the pressure * volume product is constant. You are increasing both the pressure and volume in the ram cylinder. Since the pressure vessel is a fixed volume, you are left with three possibilities: Pressure in the pressure vessel falls, or you add energy from the outside, such as externally heating the pressure vessel, or third: you have a violation of Conservation of Energy. Force that acts against the direction of motion absorbs energy, it does not supply it. Push against friction and the absorbed energy converts to heat. Push against an ideal compression spring and the spring absorbs energy and stores it. In each case the ram is performing work, and despite the fact that you do not believe it, the work is supplied from loss of internal energy in the gas volume. The only external energy that comes back into the gas is the slow heating that occurs when the temperature in the pressure vessel drops below ambient as it releases gas to the ram cylinder.
With this i agree. The compressor to the supply tank is nothing more than a battery charger is to a battery. An air compressor is one of the most inefficient devices around,and you would be lucky to turn 20% of the energy used by the compressor into gas energy.
And knowing that fact should suggest to you that your idea that you can retain all the energy inside the gas volume while delivering energy to the outside world is wrong. If your idea were correct, then we could leverage it to make over 100% efficient compressors.
Do you mean as the ram expands the gas inside cool's?. If so,this is not the case,as the pressure is always increasing until it reaches it's regulated pressure.(see original diagram at start of thread)The reason the pressure continually rises is because the return springs are progresive. This remains the case regardless of wether there is a heavier resistance placed on the ram or not.
Referring to your figure in post #49, assuming the regulator is arbitrarily fast, I am inclined to agree with you that the ram cylinder heats throughout the stroke. Pressure goes from 0psig initially and moves rapidly to 50psig and holds there by virtue of the arbitrarily fast regulator until the ram fully extends at which point gas flow falls to zero. As the ram cylinder cools the regulator will theoretically admit additional gas in order to maintain the pressure value.
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The resolution of the digital gauges i am useing is .1psi for pressure,and .1*C for temperature.
Taking the gross assumption that the resolution and absolute accuracy are one in the same, then the question becomes what are the volumes of your pressure vessel and ram cylinder?
There is another way to test such a device useing only a starting pressure and temperature-->and these are only needed to insure each test starts with the same perameters/stored energy..
Can you guess what this test may be Mark-LE ? This test is indisputable.
Sorry, but attempting to perform an energy balance for a process without knowing the starting and ending energies of the process does not compute with me.
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Taking the gross assumption that the resolution and absolute accuracy are one in the same, then the question becomes what are the volumes of your pressure vessel and ram cylinder?
Pressure vessel is 20ltr's,and ram is 750ml's-->but i am yet to liquid check them. I also need to calculate line and regulator volumes.
Sorry, but attempting to perform an energy balance for a process without knowing the starting and ending energies of the process does not compute with me.
I am supprised Mark. I would have thought you would have more than one way to check as to which test showed an increase in energy. As i stated,we know the start energy,but you dont need to have gauges to see which test can produce the most energy out for the given energy at the start. We just have to run repeated cycles of the ram,and see which test can achieve the most repeated cycles on the avaliable energy in the vessel. Lets just say that we run test 1 5 times,and we can achieve 13 cycles of the ram every run before the pressure in the vessel drop's to the 50psi,and the gas can no longer flow through the regulator. We then run test 2-5 time's,and in each run we can achieve 14.5 cycles of the ram before the pressure in the supply vessel drops to 50 psi,and the gas can no longer flow through the regulator. What if this happens-what if we can achieve more cycles when a resistance is placed on the ram than when there is no resistance placed on the ram?. What if that resistance is a generator coupled to a gear box,so as one movement of the ram turns this generator 50 revolutions-->and this generated power is captured in a cap bank that is 100f at 5 volt's. What is this 1250 joules of energy going to be concidered as?Part of the output energy,or part of the work done on the gas during the opperation of test 2?.
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Pressure vessel is 20ltr's,and ram is 750ml's-->but i am yet to liquid check them. I also need to calculate line and regulator volumes.
So again just for rough checks:
You've got 120psig in a 20 liter pressure vessel, and you have a ram that you fill from empty with 750ml @ 50psig:
(120psig + 15psi ambient) * 20 liters => (X psig + 15psi ambient) * 20 liters + 750ml @ 65psi - 750ml @ 15psi
If no energy is added or removed, X <= (2700 - 37.5)/20 - 15 <= 118.13 psig. If you are reading pressures higher than that, then something is very wrong.
I am supprised Mark. I would have thought you would have more than one way to check as to which test showed an increase in energy.
Why? They are your tests. I did not design them. As i stated,we know the start energy,but you dont need to have gauges to see which test can produce the most energy out for the given energy at the start. We just have to run repeated cycles of the ram,and see which test can achieve the most repeated cycles on the avaliable energy in the vessel. Lets just say that we run test 1 5 times,and we can achieve 13 cycles of the ram every run before the pressure in the vessel drop's to the 50psi,and the gas can no longer flow through the regulator. We then run test 2-5 time's,and in each run we can achieve 14.5 cycles of the ram before the pressure in the supply vessel drops to 50 psi,and the gas can no longer flow through the regulator. What if this happens-what if we can achieve more cycles when a resistance is placed on the ram than when there is no resistance placed on the ram?.
That would tell you that there is a gross misunderstanding in the test set-up, because at face value it would be claiming that performing work creates energy, which no one has ever previously observed. What if that resistance is a generator coupled to a gear box,so as one movement of the ram turns this generator 50 revolutions-->and this generated power is captured in a cap bank that is 100f at 5 volt's. What is this 1250 joules of energy going to be concidered as?Part of the output energy,or part of the work done on the gas during the opperation of test 2?.
If you perform work anywhere, that energy has to come from someplace. So if you pump up capacitors with 1250J, then you need to take at least 1250J from someplace else. If you are not measuring that loss, then barring a Nobel Prize headed your way, you have a measurement / accounting problem.
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So again just for rough checks:
You've got 120psig in a 20 liter pressure vessel, and you have a ram that you fill from empty with 750ml @ 50psig:
(120psig + 15psi ambient) * 20 liters => (X psig + 15psi ambient) * 20 liters + 750ml @ 65psi - 750ml @ 15psi
If no energy is added or removed, X <= (2700 - 37.5)/20 - 15 <= 118.13 psig.
So how much energy in joules did we loose from the 20 ltr supply vessel,and what do we now have in the 750ml cylider in joules of energy?-so as we can calculate the loss in total.
Cheers
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Tinman:
Congratulations on being selected the Experimenter Of The Month on Revolution Green.
Bill
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Tinman:
Congratulations on being selected the Experimenter Of The Month on Revolution Green.
Bill
Ya what :o >?
Im not seeing it anywhere.
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Ya what :o >?
Sorry, I now see that this was in August of 2013. Well, congratulations anyway...better late then never.
I stumbled across that story and assumed it was a current one.
Sorry,
Bill
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So how much energy in joules did we loose from the 20 ltr supply vessel,and what do we now have in the 750ml cylider in joules of energy?-so as we can calculate the loss in total.
Cheers
The equation predicts the maximum pressure that you should read based on conservation of energy. If you want to know Joules transferred we just need to identify the gas and then plug the constants into the equation.
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Sorry, I now see that this was in August of 2013. Well, congratulations anyway...better late then never.
I stumbled across that story and assumed it was a current one.
Sorry,
Bill
Lol-all good Bill,as i missed it to lol.
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The equation predicts the maximum pressure that you should read based on conservation of energy. If you want to know Joules transferred we just need to identify the gas and then plug the constants into the equation.
The gas is just ambiant air. I was just wondering how much energy was lost during the transfer from tank to ram. So we start with X amount of energy in the tank,and 0 in the ram. We take some from the tank,and put it in the ram. So lets say the temperatures remain the same,what energy in joules was removed from the tank,and what energy in joules was gained in the ram?
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The gas is just ambiant air. I was just wondering how much energy was lost during the transfer from tank to ram. So we start with X amount of energy in the tank,and 0 in the ram. We take some from the tank,and put it in the ram. So lets say the temperatures remain the same,what energy in joules was removed from the tank,and what energy in joules was gained in the ram?
Gamma for the nitrogen and oxygen which makes up most of the air is 1.4. So:
P*V1.4 = K.
Whereas the energy is P*V.
Consequently, for a change in volume from VSTART to VEND the remaining energy as a ratio to the starting energy is:
EEND/ESTART = PSTART*(VSTART1.4/VEND0.4)/(PSTART*VSTART)
PSTART appears in the numerator and the denominator and cancels, and VSTART is raised to the 1.4th power in the numerator and the 1.0th power in the denominator, so simplifying we get the remaining energy percentage as:
EEND/ESTART = (VSTART/VEND)0.4 and the percentage energy lost is:
EINTERNAL_LOSS/ESTART = (VEND0.4 - VSTART0.4)/VEND0.4
I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes. For the first approximation we would just use the pressure vessel volume for VSTART and the sum of the cylinder extension volume plus the pressure vessel volume for VEND.
For 20 liter pressure vessel, and a 750ml ram, the approximate percentage loss to heat on each stroke is:
ELOSS% = (20.750.4 - 200.4 )/20.750.4 = 1.46%
The initial energy is 120psi * 20 liters = 6895 P/psi * 120 psi * 0.02m3 ~= 16.5kJ. 1.46% * 1.65kJ ~= 242J internal energy loss per stroke. This is approximate because we are treating the ram cylinder volume as completely void, when it at 1 ATM. This rough approximation s/b within 5-10% of the actual loss. If you vent the ram to reset it, then after about 50 operations the energy in the pressure vessel and the pressure should both be about 50% of the starting value.
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I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes. For the first approximation we would just use the pressure vessel volume for VSTART and the sum of the cylinder extension volume plus the pressure vessel volume for VEND.
Thanks Mark
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Gamma for the nitrogen and oxygen which makes up most of the air is 1.4. So:
P*V1.4 = K.
Whereas the energy is P*V.
Consequently, for a change in volume from VSTART to VEND the remaining energy as a ratio to the starting energy is:
EEND/ESTART = PSTART*(VSTART1.4/VEND0.4)/(PSTART*VSTART)
PSTART appears in the numerator and the denominator and cancels, and VSTART is raised to the 1.4th power in the numerator and the 1.0th power in the denominator, so simplifying we get the remaining energy percentage as:
EEND/ESTART = (VSTART/VEND)0.4 and the percentage energy lost is:
EINTERNAL_LOSS/ESTART = (VEND0.4 - VSTART0.4)/VEND0.4
I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes. For the first approximation we would just use the pressure vessel volume for VSTART and the sum of the cylinder extension volume plus the pressure vessel volume for VEND.
For 20 liter pressure vessel, and a 750ml ram, the approximate percentage loss to heat on each stroke is:
ELOSS% = (20.750.4 - 200.4 )/20.750.4 = 1.46%
The initial energy is 120psi * 20 liters = 6895 P/psi * 120 psi * 0.02m3 ~= 16.5kJ. 1.46% * 1.65kJ ~= 242J internal energy loss per stroke. This is approximate because we are treating the ram cylinder volume as completely void, when it at 1 ATM. This rough approximation s/b within 5-10% of the actual loss. If you vent the ram to reset it, then after about 50 operations the energy in the pressure vessel and the pressure should both be about 50% of the starting value.
So to get this straight,each stroke we loose heat?,and to where-the enviroment?
And next-->this heat loss results in a drop of preasure.?
So you are sure there is no way we could gain energy from the enviroment through the open part of the system(other than heat)/.
I have all the equipment now,and we are going to change things around a little to make the outcome much easyer to measure. We will be adding a second tank to capture this !extra! energy.
I will draw up a schematic when i get back from my next freight trip-should have it here by saterday arvo.
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Each stroke you lose internal energy in the gas to heat that then eventually finds its way out of the system. You have practical experience with rams and know they heat up as they operate. You have practical experience with aerosol cans and know that they cool down.
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Each stroke you lose internal energy in the gas to heat that then eventually finds its way out of the system. You have practical experience with rams and know they heat up as they operate. You have practical experience with aerosol cans and know that they cool down.
The thing with the aerosol can is you have a liquid turning to gas when the gas is released from the can-same as the tap freezing on a LPG bottle when you open the valve. This rapid expansion is what creates the cooling effect. But what if the expansion is gas expanding slowly,and in the other vessel the gas is being compressed at a faster rate?.
Tomorrow i will liquid measure the two tanks,and post a diagram of how the first test setup will be.
What i will need to know after i have the tank capacities,is what pressures are needed in each tank to be a unity measurement.
So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.
The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.
More info tomorrow.
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The thing with the aerosol can is you have a liquid turning to gas when the gas is released from the can-same as the tap freezing on a LPG bottle when you open the valve. This rapid expansion is what creates the cooling effect. But what if the expansion is gas expanding slowly,and in the other vessel the gas is being compressed at a faster rate?.
The change in internal energy depends on the degrees of freedom of the molecules and the change in volume. How fast you change the volume determines the power.
Tomorrow i will liquid measure the two tanks,and post a diagram of how the first test setup will be.
What i will need to know after i have the tank capacities,is what pressures are needed in each tank to be a unity measurement.
So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.
The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.
More info tomorrow.
If you use a small tank as your reservoir raising the pressure in a larger vessel then you will lose a greater percentage of energy with each operation than if you use a large tank as the high pressure reservoir. Remember that when you calculate energy you need to use absolute temperature and pressure values.
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So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.
The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.
More info tomorrow.
This sounds very interesting!
Looking forward for the results.
Regards
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The change in internal energy depends on the degrees of freedom of the molecules and the change in volume. How fast you change the volume determines the power.If you use a small tank as your reservoir raising the pressure in a larger vessel then you will lose a greater percentage of energy with each operation than if you use a large tank as the high pressure reservoir. Remember that when you calculate energy you need to use absolute temperature and pressure values.
What im trying to ask is-if we take 2 vessels with a volume of say 5 ltr's each,and fill one of those vessels to a pressure of say 40psi,and we then open the valve so as gas flows into the empty vessel,then in an ideal situation where there is no losses,we would have 20psi in each vessel. So we have halved the pressure but doubled the volume. Dose this then mean that we still have the same amount of energy in those two vessels combined as we did at the start with the 40psi in just the one vessel?. If so,then to obtain a higher energy level than when we started,i would just have to raise the end pressure in the combined vessels by say 1psi to 21psi each. Would this then be proof enough that by making the gas do work can indeed raise the energy level of that gas..?.
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If you have two vessels each of some volume V and you pressurize one to 1 ATM and the other to 1 ATM + 40 psi with N2 and O2 then the math works as such:
ESTART = K1*V * (40+15+15) = 70*K1*V
If the energy were distributed across both vessels evenly, then each would have 35*K1*V which is the 20psi you predict.
But since: P*V1.4 = K this does not happen and you will have less than 20psi in each vessel.
To more directly answer your question: Consider that if equalizing pressure between two vessels left us with more energy than we had at the start. Then in order to operate pneumatic machinery we could just keep coupling larger and larger storage tanks to each other and then use the energy stored at reduced pressure and very high volume to do our work and recompress a new seed quantity of gas. That of course does not work. We'll see what happens in your experiment.
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If you have two vessels each of some volume V and you pressurize one to 1 ATM and the other to 1 ATM + 40 psi with N2 and O2 then the math works as such:
ESTART = K1*V * (40+15+15) = 70*K1*V
If the energy were distributed across both vessels evenly, then each would have 35*K1*V which is the 20psi you predict.
But since: P*V1.4 = K this does not happen and you will have less than 20psi in each vessel.
To more directly answer your question: Consider that if equalizing pressure between two vessels left us with more energy than we had at the start. Then in order to operate pneumatic machinery we could just keep coupling larger and larger storage tanks to each other and then use the energy stored at reduced pressure and very high volume to do our work and recompress a new seed quantity of gas.
Exactly ;)
Now,whats with the and you pressurize one to 1 ATM
There is no !pressurizing to 1 ATM,as it is already at 1 ATM. As 1 ATM in the tank is equal to the pressure of the outer enviroment,then there is a 0 potential difference,and no work can be done. Our gauges will read 0,both differential nad gauge pressure. So why say !!pressurize to 1 ATM? ,when you dont have any pressure at 1 ATM. If we have a vessel pressurized at 1 ATM,dose that vessel have any stored energy in it?-->if so,then our whole atmostphere must be 1 big storage of the same energy/
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Exactly ;)
Now,whats with the and you pressurize one to 1 ATM
There is no !pressurizing to 1 ATM,as it is already at 1 ATM. As 1 ATM in the tank is equal to the pressure of the outer enviroment,then there is a 0 potential difference,and no work can be done.
The internal energy in the gas is based on absolute pressure. You will reduce errors by always going back to absolute pressure and temperature.
Our gauges will read 0,both differential nad gauge pressure. So why say !!pressurize to 1 ATM? ,when you dont have any pressure at 1 ATM. If we have a vessel pressurized at 1 ATM,dose that vessel have any stored energy in it?-->if so,then our whole atmostphere must be 1 big storage of the same energy/
Yes the atmosphere stores a tremendous amount of energy.
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Ok,i have just messured the two tank's,and i was out by double lol.
The small tank is exactly 10 ltr's,and the big tank is exactly 20 ltr's.
There is a reason i will be using the small tank for the pressurized tank,and that will become apparent later on in the testing. The two tanks will be joined via a ball valve that will be opened to let the pressurized gas drain into the large tank. Each tank will have it's own temperature gauge,but only the small tank will need a pressure gauge-as the pressure at the end of each test will be the same in both tanks.
So a quick bit of calculating for you Mark,if you could please.
Our small tank(10LTR) will have a pressure of 40psi gauge pressure at the start of the test,and the large tank will have 1 ATM. The temperature will be close to 25*C-avaerage ambiant temperature here at the moment-->will give exact temperature on day of test. But going on these measurement's,can you calculate what we should have in the way of temperature in each tank,and the pressure in both tanks?.
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Ok,i have just messured the two tank's,and i was out by double lol.
The small tank is exactly 10 ltr's,and the big tank is exactly 20 ltr's.
There is a reason i will be using the small tank for the pressurized tank,and that will become apparent later on in the testing. The two tanks will be joined via a ball valve that will be opened to let the pressurized gas drain into the large tank. Each tank will have it's own temperature gauge,but only the small tank will need a pressure gauge-as the pressure at the end of each test will be the same in both tanks.
So a quick bit of calculating for you Mark,if you could please.
Our small tank(10LTR) will have a pressure of 40psi gauge pressure at the start of the test,and the large tank will have 1 ATM. The temperature will be close to 25*C-avaerage ambiant temperature here at the moment-->will give exact temperature on day of test. But going on these measurement's,can you calculate what we should have in the way of temperature in each tank,and the pressure in both tanks?.
Roughly:
55 psi * 101.4 liters +
15 psi * 201.4 liters
= PEQUALIZED * 301.4 liters
PEQUALIZED = ((55*25.1) + (15*66.3))/116.9 = 20.3 psi = 5.3psi gauge
Applying the ideal gas law: PV = nRT, T = PV/(nR). TEND/TSTART = (PV)END/(PV)START * nSTART/nEND
The small tank PV product at the end K*20.3psi*10l instead of K*55psi*10l.
So (PV)END/(PV)START = 20.3/55 = 0.369.
The proportion of molecules in the small tank changes from 55*10/(55*10 + 15*20) to 10/30.
So, nSTART/nEND = 550/850 * 3/1 = 1.941.
The absolute temperature then is: TEND = 0.369 * 1.941 * TSTART = 0.716 * TSTART.
If TSTART = 25C = 298.2K then TEND = 214K = -59C.
The large tank PV product at the end K*20.3psi*20l instead of K*15psi*20l.
So (PV)END/(PV)START = 406/300 = 1.353.
The proportion of molecules in the large tank changes from 15*20/(55*10 + 15*20) to 20/30.
So, nSTART/nEND = 300/850 * 3/2 = 0.529.
The absolute temperature then is: TEND = 1.353 * 0.529 * TSTART = 0.716 * TSTART.
If TSTART = 25C = 298.2K then TEND = 214K = -59C.
I am a bit skeptical that the predicted equilibrium temperature is so low, but the work is there to see.
Anyway, this gives us a large separation from what constant energy predicts, which is that you would get ((10*(15+40)) + (20*15))/30 - 15 = 10*40/30 = 13.3psi gauge. And since under the constant energy premise: the PV product is constant as is the number of molecules there would be no temperature change.
So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.
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So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.
I'd predict you will see the gauge pressure of 13.3 psi fairly quickly. What will happen is heat will flow from ambient surroundings into the combined cylinders as the gas expansion causes them to become cooler. The rate of change of temperature can't be calculated without knowing how well insulated they are and would be described by differential equations.
You would only see the lower pressure value if the cylinders are perfectly insulated.
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I'd predict you will see the gauge pressure of 13.3 psi fairly quickly. What will happen is heat will flow from ambient surroundings into the combined cylinders as the gas expansion causes them to become cooler. The rate of change of temperature can't be calculated without knowing how well insulated they are and would be described by differential equations.
You would only see the lower pressure value if the cylinders are perfectly insulated.
Because the tanks are pretty reflective most of the transfer will be conductive. The thermal resistance of the gas is pretty high. I expect that the time constant is at least a couple minutes. That's plenty of time for tinman to take his pressure and temperature readings. Even as the system moves on him, he can get a decent qualitative result. Any pressure well below 13.3psi and any temperature well below ambient should be adequate proof to him that the PV product is not constant. If he wants he can later pack the whole thing in some decent insulation to show himself how that slows down the return to room temperature.
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Roughly:
I am a bit skeptical that the predicted equilibrium temperature is so low, but the work is there to see.
Anyway, this gives us a large separation from what constant energy predicts, which is that you would get ((10*(15+40)) + (20*15))/30 - 15 = 10*40/30 = 13.3psi gauge. And since under the constant energy premise: the PV product is constant as is the number of molecules there would be no temperature change.
So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.
So the first thing i need to do is run say 5(what we will call) open test's. This will give us a figure to work with,and one to beat. These test will be just opening the ball valve between the two tank's,and let the gas flow from one to another until an equilibrium is reached between the two tank's. If i record the start pressure in the small tank(the other will be 1 ATM of corse),and the temperature in both(as well as ambiant temperature),then open the valve to allow gas flow,then record the end pressure in both tanks-along with end temperature in both tank's as soon as gas flow has stopped,then this will give us close to an instantaneous measurement. I will then leave the system rest for a period of 2 minutes,and once again record the values. This will give us an indication as to how much the enviroment is contributing in the way of heat energy within that 2 minutes.
When i apply my first system,we will carry out the tests again as we did above. With the first system in opperation,it will take about 2 minutes to complete one cycle-->pressure in both tanks reach equilibrium.
Regardless of what LE says about drawing heat in from the enviroment,the heat of the gas can never exceed that of the outside enviromental temperature-->if it dose,we then have free energy.
The goal is to raise the pressure higher than the 13.3psi Mark has stated. But first we must see if that is indeed what we will have when the primary test are carried out on the system.
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So the first thing i need to do is run say 5(what we will call) open test's. This will give us a figure to work with,and one to beat. These test will be just opening the ball valve between the two tank's,and let the gas flow from one to another until an equilibrium is reached between the two tank's. If i record the start pressure in the small tank(the other will be 1 ATM of corse),and the temperature in both(as well as ambiant temperature),then open the valve to allow gas flow,then record the end pressure in both tanks-along with end temperature in both tank's as soon as gas flow has stopped,then this will give us close to an instantaneous measurement. I will then leave the system rest for a period of 2 minutes,and once again record the values. This will give us an indication as to how much the enviroment is contributing in the way of heat energy within that 2 minutes.
When i apply my first system,we will carry out the tests again as we did above. With the first system in opperation,it will take about 2 minutes to complete one cycle-->pressure in both tanks reach equilibrium.
Regardless of what LE says about drawing heat in from the enviroment,the heat of the gas can never exceed that of the outside enviromental temperature-->if it dose,we then have free energy.
The goal is to raise the pressure higher than the 13.3psi Mark has stated. But first we must see if that is indeed what we will have when the primary test are carried out on the system.
LibreEnergie is not suggesting that the environment is adding energy to the tanks over and above what was originally there. When the valve opens and the gasses are moving the small more pressurized tank temperature drops, while the bigger tank at 1 ATM heats rejecting heat to the environment. If you put the whole thing in one really well insulated container then as the tanks equalize, their temperatures will also equalize getting much colder in the process. It takes heat from outside to restore the total energy seen as the sum total PV of the combined tanks.
The longer it takes the tanks to equalize the less sever the temperature and pressure drop will be. By the seat of my pants a couple of minutes is probably short enough that you will see a significant temperature drop and pressure shortfall versus the constant energy / constant PV produce hypothesis that you are testing.
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LibreEnergie is not suggesting that the environment is adding energy to the tanks over and above what was originally there. When the valve opens and the gasses are moving the small more pressurized tank temperature drops, while the bigger tank at 1 ATM heats rejecting heat to the environment. If you put the whole thing in one really well insulated container then as the tanks equalize, their temperatures will also equalize getting much colder in the process. It takes heat from outside to restore the total energy seen as the sum total PV of the combined tanks.
The longer it takes the tanks to equalize the less sever the temperature and pressure drop will be. By the seat of my pants a couple of minutes is probably short enough that you will see a significant temperature drop and pressure shortfall versus the constant energy / constant PV produce hypothesis that you are testing.
We will call the small tank(10ltr) tank A,and the larger tank(20ltr)tank B from here on in.
So to sum it up,if we pressurise tank A to 40psi gauge pressure,and release that pressurised gas into tank B,to be an equil amount of energy now in both tanks that equils the energy we had in tank A to start with,the pressure in each tank would have to be 13.3psi-->is this correct?.
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We will call the small tank(10ltr) tank A,and the larger tank(20ltr)tank B from here on in.
So to sum it up,if we pressurise tank A to 40psi gauge pressure,and release that pressurised gas into tank B,to be an equil amount of energy now in both tanks that equils the energy we had in tank A to start with,the pressure in each tank would have to be 13.3psi-->is this correct?.
13.3psi gauge and constant temperature is what would result without any energy loss. Constant energy is what you have claimed, and LE and I dispute.
After equalization of temperature and pressure the energy in each tank is proportional to its respective volume. E = PV.
Applying textbook thermodynamics: Were the system to equalize instantly the A tank pressure would fall to 5.3psi gauge while the B tank pressure rises to 5.3psi gauge and the temperature in each would if allowed to equalize while perfectly insulated from the outside world at -59C. It won't equalize instantly so the pressure will be between 5.3psi gauge and 13.3 gauge, and the temperature will bottom out between -59C and the starting stabilized temperature. Once the pressure and temperature bottom out, the environment will reheat the contents so that if you come back say an hour later the temperature will be very close to the outside air temperature and the pressure will be very close to 13.3psi.
If you were to only partially equalize then the A tank would cool and the B tank would heat during each transfer.
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13.3psi gauge and constant temperature is what would result without any energy loss. Constant energy is what you have claimed, and LE and I dispute.
After equalization of temperature and pressure the energy in each tank is proportional to its respective volume. E = PV.
Applying textbook thermodynamics: Were the system to equalize instantly the A tank pressure would fall to 5.3psi gauge while the B tank pressure rises to 5.3psi gauge and the temperature in each would if allowed to equalize while perfectly insulated from the outside world at -59C. It won't equalize instantly so the pressure will be between 5.3psi gauge and 13.3 gauge, and the temperature will bottom out between -59C and the starting stabilized temperature. Once the pressure and temperature bottom out, the environment will reheat the contents so that if you come back say an hour later the temperature will be very close to the outside air temperature and the pressure will be very close to 13.3psi.
If you were to only partially equalize then the A tank would cool and the B tank would heat during each transfer.
Well i have carried out over ten test now,and the results are always within 2 decimal points,both in temperature and pressure. Never once did i see a tem drop any where near -59*C-->something is amiss there ???
I ran 5 calibrated test,and averaged out the results over the 5 test. As i stated,all ten test carried out only seen a .2 swing either way in temp and pressure.
The results are as below.
Tank A-1 ATM-temp 22*C-->that is also the ambiant temperature at the time of the test's.
Tank B-1 ATM-temp 22*C
Tank A pressurised to 40 psi gauge pressure-temp 29*C
Valve opened to tank B ,and aloud to equalise.
Pressure tank's A&B=13.5psi
Temp tank A=19*C
Temp tank B=31*C
Left for 2 minutes.
Pressure tank's A&B-13.5psi
Temp tank A=22*C
Temp tank B=24*C
The added plumbing has added 65mL to tank A,and 50mL to tank B
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Well i have carried out over ten test now,and the results are always within 2 decimal points,both in temperature and pressure. Never once did i see a tem drop any where near -59*C-->something is amiss there ???
I ran 5 calibrated test,and averaged out the results over the 5 test. As i stated,all ten test carried out only seen a .2 swing either way in temp and pressure.
The results are as below.
Tank A-1 ATM-temp 22*C-->that is also the ambiant temperature at the time of the test's.
Tank B-1 ATM-temp 22*C
Tank A pressurised to 40 psi gauge pressure-temp 29*C
Valve opened to tank B ,and aloud to equalise.
Pressure tank's A&B=13.5psi
Temp tank A=19*C
Temp tank B=31*C
Left for 2 minutes.
Pressure tank's A&B-13.5psi
Temp tank A=22*C
Temp tank B=24*C
The added plumbing has added 65mL to tank A,and 50mL to tank B
A couple of comments and questions:
How have you fixtured your temperature sensors? Are they suspended in the tanks or are they in contact with the tank walls?
How long does it take your tanks to equalize?
How heavy are your tanks proper?
Did you let tank A come back down to your 22 ambient before connecting it to tank B, or did you connect it while it was still warm? You should find that if you fill tank A to 40 psi gauge and then let it sit that the pressure drifts down and it will take a series of shorter and shorter top-ups from your pressurized air source to get tank A to 40 psi gauge and keep it there after it is allowed to sit.
The indication of pressure at 13.5psi gauge suggests that enough heat is coming back from the tank walls to reheat the gas.
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A couple of comments and questions:
You should find that if you fill tank A to 40 psi gauge and then let it sit that the pressure drifts down and it will take a series of shorter and shorter top-ups from your pressurized air source to get tank A to 40 psi gauge and keep it there after it is allowed to sit.
How have you fixtured your temperature sensors? Are they suspended in the tanks or are they in contact with the tank walls?
The temp sensors are suspended in the tanks ,so as they in the center of the tanks.
How long does it take your tanks to equalize?
I can controll the flow rate,and tried a few different times,from instant(within a second),and up to 30 seconds. There didnt seem to be much difference.
How heavy are your tanks proper?
Tanks proper?. Im guessing here,but do you mean mounting bracket's?. If so,there isnt any,the tanks are just sitting on the wodden bench ATM.
Did you let tank A come back down to your 22 ambient before connecting it to tank B, or did you connect it while it was still warm?
They were conected while still warm. I will carry out those test after dinner tonight-->letting the temperature drop back down to ambiant before opening the valve to tank B.
The indication of pressure at 13.5psi gauge suggests that enough heat is coming back from the tank walls to reheat the gas.
This is a good thing for our setup to acheive it's goal.
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Thanks. I want to know the weight of the tanks themselves empty. I assume they are steel. This would give me a good indication of their thermal capacity. The two things that I have not taken into account yet is the amount of heat that the tanks can store in their walls and Van der Waals forces.
If for even a rapid: say 5 second or 10 second equalization you are getting back to a pressure indicative of constant energy then the tanks are recuperating for you as big heat sinks / sources. The good, bad, and indifferent about that is that if your target machine has similar geometries and you operate at a low rate then you won't end up losing much energy just to moving gas from your main vessel. The primary energy that you lose from within the combined pressure vessel and ram will be the work the ram performs on the outside world.
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Thanks. I want to know the weight of the tanks themselves empty. I assume they are steel. This would give me a good indication of their thermal capacity. The two things that I have not taken into account yet is the amount of heat that the tanks can store in their walls and Van der Waals forces.
If for even a rapid: say 5 second or 10 second equalization you are getting back to a pressure indicative of constant energy then the tanks are recuperating for you as big heat sinks / sources. The good, bad, and indifferent about that is that if your target machine has similar geometries and you operate at a low rate then you won't end up losing much energy just to moving gas from your main vessel. The primary energy that you lose from within the combined pressure vessel and ram will be the work the ram performs on the outside world.
I will have to disconect the plumbing to weigh the tank's. I will do this tomorrow when i make the carry frame for the tanks. The mounting brackets have a rubber band that go's between them and the tanks,so that should provide some insulating properties between tank and mounting brackets. The brackets are also only an inch wide,so little surface contact anyway.
OK,the results of the tests you asked for Mark.
I filled tank A with 45psi in the hope that when it dropped down to ambiant temperature,it would be close to 40psi. Ambiant temperature at the time of test was 20*C,and it took about 5 minutes for the temperature of the gas in tank A to drop to that temperature. The pressure ended up at 40.4psi-so was preaty close.
Starting pressure tank A 40.4psi gauge pressure.
Temperature of both tabks was 20*C
Tank B=1 ATM.
End result was
Tanks A&B end pressure= 13.1psi
Tank A instant end temperature was 17*C
Tank B instant end temperature was 23*C
After 5 minutes sitting.
Tanks A&B pressure still 13.1psi
End temperature for both tanks was of course 20*C
I would also like you to crunch these numbers for me if you could Mark.
These are the average results over 4 runs with my first system enabled
Start of test was as above.
Tank A start pressure = 40.1psi
Tank A&B start temperature was 20*C.
End results.
Tank A&B end pressure = 16.2 psi
Tank A&B end temperature is of course 20*C.<--- This is NOT the ram test results. This is a much simpler design.
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I will have to disconect the plumbing to weigh the tank's. I will do this tomorrow when i make the carry frame for the tanks. The mounting brackets have a rubber band that go's between them and the tanks,so that should provide some insulating properties between tank and mounting brackets. The brackets are also only an inch wide,so little surface contact anyway.
OK,the results of the tests you asked for Mark.
I filled tank A with 45psi in the hope that when it dropped down to ambiant temperature,it would be close to 40psi. Ambiant temperature at the time of test was 20*C,and it took about 5 minutes for the temperature of the gas in tank A to drop to that temperature. The pressure ended up at 40.4psi-so was preaty close.
Starting pressure tank A 40.4psi gauge pressure.
Temperature of both tabks was 20*C
Tank B=1 ATM.
End result was
Tanks A&B end pressure= 13.1psi
Tank A instant end temperature was 17*C
Tank B instant end temperature was 23*C
After 5 minutes sitting.
Tanks A&B pressure still 13.1psi
End temperature for both tanks was of course 20*C
I would also like you to crunch these numbers for me if you could Mark.
These are the average results over 4 runs with my first system enabled
Start of test was as above.
Tank A start pressure = 40.1psi
Tank A&B start temperature was 20*C.
End results.
Tank A&B end pressure = 16.2 psi
Tank A&B end temperature is of course 20*C.<--- This is NOT the ram test results. This is a much simpler design.
ATM is assumed 101.3E3P
Start
Tank A .01m3 278.6E3P gauge 379.9E3P abs 293.2K
Tank B .02m3 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.9
End
Tank A .01m3 90.3E3P gauge 191.6E3P abs 290.2K
Tank B .02m3 90.3E3P gauge 191.6E3P abs 296.2K
sum(PV/T) = 19.5
Energy instant end/Energy instant start = 19.5/19.9 = 98%
The second test data does not make sense. It suggests accuracy issues in the measurements.
Start
Tank A .01m3 276.5E3P gauge 377.8E3P abs 293.2K
Tank B .02m3 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.8
End
Tank A .01m3 111.7E3P gauge 213.0E3P abs 293.2K
Tank B .02m3 111.7E3P gauge 213.0E3P abs 293.2K
sum(PV/T) = 21.8
Energy instant end/Energy instant start = 21.8/19.8 = 110%
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Energy instant end/Energy instant start = 19.5/19.9 = 98%
The second test data does not make sense. It suggests accuracy issues in the measurements.
Start
Tank A .01m3 276.5E3P gauge 377.8E3P abs 293.2K
Tank B .02m3 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.8
End
Tank A .01m3 111.7E3P gauge 213.0E3P abs 293.2K
Tank B .02m3 111.7E3P gauge 213.0E3P abs 293.2K
sum(PV/T) = 21.8
Energy instant end/Energy instant start = 21.8/19.8 = 110%
Do you find it odd that it is assumed that the tests were carried out correctly when an under 100% figure was achieved,and yet when an over 100% figure was achieved,then there must have been measurement error-->even though the tests were carried out in the very same way,with the very same equipment,in identical conditions.
All one has to do Mark to achieve this,is to introduce more gas(air) from the enviroment-->the open part of the system. You simply use the energy avaliable in tank A to pull in more gas from the surroundings of the device. If you think hard enough,you will be able to work out how you can draw more gas into the system without letting any of the gas contained within the system escape. Think along the lines of an gas diode,and you are on the right track. You will also then be able to work out why i chose the small tank as the energy storage tank,and the large tank as the collector. By using a very small jet,we can achieve a high speed air flow at low flow rates. It is this that is needed to achieve the effect you have seen in the results i posted. You must agree that even the smallest amount of gas introduced into the system from the enviroment,will only add to the end resulting energy amount.
Im not sure if you are able to do this calculation,but can you derive as to how much extra gas would be needed to be introduced to the system to achieve a result of 110%-eg-1 ltr at ATM ?.
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Do you find it odd that it is assumed that the tests were carried out correctly when an under 100% figure was achieved,and yet when an over 100% figure was achieved,then there must have been measurement error-->even though the tests were carried out in the very same way,with the very same equipment,in identical conditions.
Test results that at least superficially agree with 200+ years of thermodynamics pass sanity checks. Tests that don't have that 200+ year hill of experience to overcome.
All one has to do Mark to achieve this,is to introduce more gas(air) from the enviroment-->the open part of the system. You simply use the energy avaliable in tank A to pull in more gas from the surroundings of the device. If you think hard enough,you will be able to work out how you can draw more gas into the system without letting any of the gas contained within the system escape. Think along the lines of an gas diode,and you are on the right track. You will also then be able to work out why i chose the small tank as the energy storage tank,and the large tank as the collector. By using a very small jet,we can achieve a high speed air flow at low flow rates. It is this that is needed to achieve the effect you have seen in the results i posted. You must agree that even the smallest amount of gas introduced into the system from the enviroment,will only add to the end resulting energy amount.
If you add energy by doing work then what is special? At this point the most basic part of the rig is still being shaken out. If the tests are as represented then the second set data is not reasonable. If there is something you have not disclosed about the second set of tests, then having me run calculations on faulty presumptions is bound to yield GIGO results.
Im not sure if you are able to do this calculation,but can you derive as to how much extra gas would be needed to be introduced to the system to achieve a result of 110%-eg-1 ltr at ATM ?.
The question does not make any sense to me in the context of your two fixed volume closed container set-up.
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Wow
It's aint "all over" yet but it's shure has been intresting.
Nice work guys (all)
best wishes (all)
floor
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Test results that at least superficially agree with 200+ years of thermodynamics pass sanity checks. Tests that don't have that 200+ year hill of experience to overcome.
Unlike most,i would have to agree with you on this
If you add energy by doing work then what is special?
Because the work being done increases the net energy left after each cycle,as can be seen with the test results i supplied.
At this point the most basic part of the rig is still being shaken out.
Yes,i agree. And today i will be mounting everything solid so as we can obtain more accurate readings in a shorter time.I will also be going on the hunt for a second pressure gauge,so as we can check one against the other. I also noticed after doing a leak test with soapy water,that i do have a small leak on one of the bung's-->this would have actually been going against us a little.
If the tests are as represented then the second set data is not reasonable.
The only reason the second test is assumed !not reasonable! is because it go's against every thing you believe,and not because the suppllied data is incorrect.
If there is something you have not disclosed about the second set of tests, then having me run calculations on faulty presumptions is bound to yield GIGO results.
I believe i have said many times that extra energy can be drawn in from the open part of the system-the enviroment which surrounds the system. the first set of tests were to get a calibration of the system,and as you seen Mark,we were at 90% efficiency,and i believe that if the small leak is repaired,then the system will come in at 100%. This means the system is accurate,and will provide accurate test data,and gives us a solid test bed to work with. Im not much good at all those mathmatical numbers ,so that is why i have asked you to crunch them for me.I provide you with the data,and you supply us with the calculations.--> i must say at this point in time that i do appreciate your time. ;)
The question does not make any sense to me in the context of your two fixed volume closed container set-up.
Yes,the two volumes of the tanks are fixed,but the end volume of gas increases as the transfer takes place. Here we have a case where the energy within tank A can do work without any energy loss,and the net result is an energy gain. To put it in a different way,we have a battery providing energy to a system,but the energy within the system climbs to a higher rate to that of which the battery is supplying.
The thing that makes this work,and is also a known and proven fact,is-->a gas flowing at high velocity can create a vacuum. This vacuum is what causes an increase in the volume of gas within the system at the end of each cycle.
I think Webby is catching on,but i hope he holds off until i have the system completed with the ram all setup and tested.
Quote:
That is an interesting approach you are describing,, now I think I finally understand WHY you want to use a ram,, tricky
Below is a quick sketch of the system that achieved the results provided in the last test that makes no sense Mark. Like i said,only energy from the open system(the surrounding enviroment)can enter the system,the stored energy within the system cannot leave the system. This means a net gain in energy. The sketch is just a proof of concept,and dose not relate to my exact design of the venturi i use,as that is one of my own designs,and will be staying with me for the time being.
https://www.youtube.com/watch?v=Wokswr_KHXQ
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In the case of the rapid opening of the ball valve, the process being tested here is closest to adiabatic free expansion. In the idealized case the expanding gas does no work and energy in the system is constant.
P1V1 = P2V2 with constant temperature, as you have measured is the expected result.
It's not a result of any use from an over unity perspective as no external work performed.
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In the case of the rapid opening of the ball valve, the process being tested here is closest to adiabatic free expansion. In the idealized case the expanding gas does no work and energy in the system is constant.
P1V1 = P2V2 with constant temperature, as you have measured is the expected result.
It's not a result of any use from an over unity perspective as no external work performed.
The word !overunity! is a nonsense word,and used by those who do not understand the source. In my last test,i show that work has indeed been done by the flowing gas in the way of drawing in excess energy from the enviroment. The end result is that we end up with a higher energy state than we started with within the two vessel's that can perform usful work. Done the way i have done it shows that the pressurised gas in tank A can do work (via the venturi effect)without loosing any of it's energy-->in fact,the opposite applies,in that the energy state of the gas increases because the mass of gas increases. The venturi will continue to draw in gas from the enviroment until tank B reaches a pressure of 5.2psi. The check valve then closes so as no gas can escape the system. At the end of the cycle,we have an increase in gas volume,and thus,an increase in pressure which results in a higher energy state to that of which we started with.
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The word !overunity! is a nonsense word,and used by those who do not understand the source. In my last test,i show that work has indeed been done by the flowing gas in the way of drawing in excess energy from the enviroment. The end result is that we end up with a higher energy state than we started with within the two vessel's that can perform usful work. Done the way i have done it shows that the pressurised gas in tank A can do work (via the venturi effect)without loosing any of it's energy-->in fact,the opposite applies,in that the energy state of the gas increases because the mass of gas increases. The venturi will continue to draw in gas from the enviroment until tank B reaches a pressure of 5.2psi. The check valve then closes so as no gas can escape the system. At the end of the cycle,we have an increase in gas volume,and thus,an increase in pressure which results in a higher energy state to that of which we started with.
Here is the problem: You have redistributed energy such that some energy from the surrounding atmosphere ends up stored in the pair of tanks through a one time operation. The entropy both inside the tanks and outside increased in the operation. Absent performing new work you cannot restore the system to its prior state, even though you have not performed any useful work.
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The word !overunity! is a nonsense word,and used by those who do not understand the source. In my last test,i show that work has indeed been done by the flowing gas in the way of drawing in excess energy from the enviroment. The end result is that we end up with a higher energy state than we started with within the two vessel's that can perform usful work. Done the way i have done it shows that the pressurised gas in tank A can do work (via the venturi effect)without loosing any of it's energy-->in fact,the opposite applies,in that the energy state of the gas increases because the mass of gas increases. The venturi will continue to draw in gas from the enviroment until tank B reaches a pressure of 5.2psi. The check valve then closes so as no gas can escape the system. At the end of the cycle,we have an increase in gas volume,and thus,an increase in pressure which results in a higher energy state to that of which we started with.
Your claims would mean either a first or second law violation and that to me describes what most people mean as 'over-unity'.
This experiment follows well established thermodynamic principles. you won't find any 'excess energy' that is available to do work.
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Your claims would mean either a first or second law violation and that to me describes what most people mean as 'over-unity'.
This experiment follows well established thermodynamic principles. you won't find any 'excess energy' that is available to do work.
We have broken no laws at all, and the process used to increase the end energy avaliable stored in both tanks was done using an age old proven effect. There are facts that cannot be disputed here. The fact that the gas mass was increased in the system as a whole proves that it works.
If you add gas mass to the system during gas transfer, then tank 1s pressure dose not drop as low as it would otherwise. This means that the temperature also dose not drop as low as it would otherwise.
Tank Bs pressure also is raised to that of what it would be otherwise. This also means a higher end temperature.
Im sorry LE, but these facts are undisputable.
In my simple version, I have raised the avaliable energy by 10%, without breaking any laws.
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We have broken no laws at all, and the process used to increase the end energy avaliable stored in both tanks was done using an age old proven effect. There are facts that cannot be disputed here. The fact that the gas mass was increased in the system as a whole proves that it works.
If you add gas mass to the system during gas transfer, then tank 1s pressure dose not drop as low as it would otherwise. This means that the temperature also dose not drop as low as it would otherwise.
Tank Bs pressure also is raised to that of what it would be otherwise. This also means a higher end temperature.
Im sorry LE, but these facts are undisputable.
In my simple version, I have raised the avaliable energy by 10%, without breaking any laws.
It's true you haven't broken any laws. LE is saying and I agree that the premise you started with: That you are going to get free work from your ram courtesy of the surrounding atmosphere will not fly.
What you have demonstrated so far is that you can relax the system and in the course of doing so transfer some energy from the surrounding atmosphere into the tanks. If you want to show that you can leverage the underlying behaviors to cyclically perform useful work at no energy cost to the "closed system", then you need to both: Get back to your starting state, and perform some useful work that exceeds the work that you apply getting back to your starting state.
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The word !overunity! is a nonsense word,and used by those who do not understand the source. In my last test,i show that work has indeed been done by the flowing gas in the way of drawing in excess energy from the enviroment. The end result is that we end up with a higher energy state than we started with within the two vessel's that can perform usful work. Done the way i have done it shows that the pressurised gas in tank A can do work (via the venturi effect)without loosing any of it's energy-->in fact,the opposite applies,in that the energy state of the gas increases because the mass of gas increases. The venturi will continue to draw in gas from the enviroment until tank B reaches a pressure of 5.2psi. The check valve then closes so as no gas can escape the system. At the end of the cycle,we have an increase in gas volume,and thus,an increase in pressure which results in a higher energy state to that of which we started with.
If you look at your set up you will see it is not dissimilar to the operation of a heat pump. Dropping the pressure of a working fluid in the venturi allows energy to enter the system.
However, can you then use that 'extra' energy? To extract net useful work from it you HAVE to be able to cycle the system back to its original state. You'll find the 2nd law unchallenged here. The amount of energy required to do that is always more than the energy that entered the system.
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What if Tinman placed a small turbine (not unlike the type used on your faucet that lights leds) between the tanks? Then, as the air transfers, it will perform some work? The air is flowing anyway and I doubt that this small device would inhibit that flow to any appreciable amount.
Just thinking out loud here.
Bill
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What if Tinman placed a small turbine (not unlike the type used on your faucet that lights leds) between the tanks? Then, as the air transfers, it will perform some work? The air is flowing anyway and I doubt that this small device would inhibit that flow to any appreciable amount.
Just thinking out loud here.
Bill
The stored energy in the tank would either drive the turbine and produce work or allow external energy to enter by reducing the pressure through the venturi, or some combination of the two. Doing work in the turbine would reduce the pressure drop and thus the amount of external energy entering.
It's a Zero sum game.
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What if Tinman placed a small turbine (not unlike the type used on your faucet that lights leds) between the tanks? Then, as the air transfers, it will perform some work? The air is flowing anyway and I doubt that this small device would inhibit that flow to any appreciable amount.
Just thinking out loud here.
Bill
That is not unlike his original idea to get the ram to perform some external mechanical work. The internal energy of the gas transfers to the outside work done, increasing the work that must be done to return the machine to its original state by at least the same amount as the external work performed. There is a reason that the phrase: "There is no such thing as a free lunch." got coined.
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That is not unlike his original idea to get the ram to perform some external mechanical work. The internal energy of the gas transfers to the outside work done, increasing the work that must be done to return the machine to its original state by at least the same amount as the external work performed. There is a reason that the phrase: "There is no such thing as a free lunch." got coined.
I am sure you are right. I was just thinking that no air would leak outside his system with that turbine...right? I mean, it is a pass through type so the air would all end up in the 2nd tank...right? The resistance caused by the tiny turbine would increase the time the air transfer would take place a little bit but, I was considering that since all of the air would end up in the 2nd tank anyway...I didn't see where the loss would be.
Of course, if you are extracting work from a system then the energy comes from somewhere for sure, no question. I just did not see where that energy would "go" since all of the air is still transferred. Unless....that small resistance of the turbine causes "some" of the air to remain in the first tank, when it gets down close to equilibrium?
Bill
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Tinman may suffer a similar mental block. But this is fundamentally how hot gas engines work by exchanging heat and mechanical work.
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Tinman may suffer a similar mental block. But this is fundamentally how hot gas engines work by exchanging heat and mechanical work.
I see we are back to the insult's Mark,and here i thought you were starting to actually become human again-->my mistake.
Pirate,please believe me when i say it is not i that has a mental block. The block is actually coming from those that are blinded by the light!you could say. These are those that have no room for change,and assume that so called laws are set in stone,and apply to all systems in all configuration's. But you always see that when there numbers and laws dont start to add up,they revert to insult's. You see them say over and over that no work has been done,and yet the number's that MarkE crunched him self came up with a net energy gain of 10%-->and yes,work was done in the process. It take energy(work) to accelerate a mass-->gas has mas-->and the gas was accelerated through the venturi tube from a static ambiant point-->that is work being done by the gas(energy) stored in tank A. And did we loose energy when this work was done?,NO,we gained 10%. I could place a pile of beach sand below the venturi inlet,and it would suck that sand up,and displace it in tank B. Now,sand has a lot more mass per volume than air,so more work would be done-->but this still would not be satisfactory to the !laws of physics guru's.
Well Mark,would you care to entertain this mentally blocked person one more time-->just for laugh's ;D,in the way of some calculation's.
Tank A starting pressure=40.2psi
Tank A starting temperature=22*C-->this is ambiant temperature,as tank was let sit for 10 minutes this time.
Tank B starting pressure= 1 ATM
Tank B starting temperature of course 22*C
System was let sit for 20 minutes after the transfer and equalisation.
End pressure tanks A&B= 18.3psi
End temperature tanks A&B = 22*C-->ambiant temperature.
Oh,and just for the record Mark,this is the system now opperating with the ram. The work being done is the ram raising a 7kg steel plate by 400mm-->just so as you can have a laugh at the mentally blocked person on the other end of this keyboard.
Have a nice day ;)
Brad
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I see we are back to the insult's Mark,and here i thought you were starting to actually become human again-->my mistake.
Really? You think that was an insult? Really?
Pirate,please believe me when i say it is not i that has a mental block. The block is actually coming from those that are blinded by the light!you could say. These are those that have no room for change,and assume that so called laws are set in stone,and apply to all systems in all configuration's. But you always see that when there numbers and laws dont start to add up,they revert to insult's. You see them say over and over that no work has been done,and yet the number's that MarkE crunched him self came up with a net energy gain of 10%-->and yes,work was done in the process. It take energy(work) to accelerate a mass-->gas has mas-->and the gas was accelerated through the venturi tube from a static ambiant point-->that is work being done by the gas(energy) stored in tank A. And did we loose energy when this work was done?,NO,we gained 10%. I could place a pile of beach sand below the venturi inlet,and it would suck that sand up,and displace it in tank B. Now,sand has a lot more mass per volume than air,so more work would be done-->but this still would not be satisfactory to the !laws of physics guru's.
Well Mark,would you care to entertain this mentally blocked person one more time-->just for laugh's ;D,in the way of some calculation's.
When you ask for calculations against set-ups that are not as you represent them, the results are GIGO. You are free at any time to try and get your arrangements to do useful work in excess of the work that you put into them. Let me know when you think you have done so, describe the set-up accurately and I will be happy to check your work.
Tank A starting pressure=40.2psi
Tank A starting temperature=22*C-->this is ambiant temperature,as tank was let sit for 10 minutes this time.
Tank B starting pressure= 1 ATM
Tank B starting temperature of course 22*C
System was let sit for 20 minutes after the transfer and equalisation.
End pressure tanks A&B= 18.3psi
End temperature tanks A&B = 22*C-->ambiant temperature.
Oh,and just for the record Mark,this is the system now opperating with the ram. The work being done is the ram raising a 7kg steel plate by 400mm-->just so as you can have a laugh at the mentally blocked person on the other end of this keyboard.
Have a nice day ;)
Brad
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Really? You think that was an insult? Really?
When you ask for calculations against set-ups that are not as you represent them, the results are GIGO.
The device is set up to do exactly was i have claimed it would do from the very start,and that is- draw in extra energy from the open part of the system(the enviroment) while increasing in energy.
You are free at any time to try and get your arrangements to do useful work in excess of the work that you put into them.
I gave you the numbers above,and also the work that was done. The only reason that you do not wish to crunch those numbers is because the results will go against what you believe in.
Let me know when you think you have done so, describe the set-up accurately and I will be happy to check your work.
The setup is as it was first described to be,with the addition of tank B.Tank A supplying a ram with compressed air. The ram doing usful work(lifting a 7kg weight 400mm),and the(now heated)gas returned to tank B. The rest of the opperation will be remaining with me pending your number crunching-If you will please :). Once i have the system to 150% by the numbers i supply you,i will then go about finding a way to present the system here on this thread so as it cannot be refuted. This will be done in a way that is to satisfy !your! guidlines Mark-->which i have tried to follow to the leter to this date.
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The device is set up to do exactly was i have claimed it would do from the very start,and that is- draw in extra energy from the open part of the system(the enviroment) while increasing in energy.
Of course you can draw in extra energy from the environment by using some or all of the stored potential the cylinder at 40 psi. Then what?
To be useful you have to be able to recompress that cylinder without using any FURTHER external work (remember , initially you did some work to raise the pressure in the tank, but conveniently ignore this in your specification of the 'system').
The best you will be able to achieve is that you can extract energy from the system equal to the amount of work performed to pressurise it initially, such as would occur during a reversible adiabatic process. The amount you can extract as useful work compared with the amount lost as heat will depend on the thermodynamic cycle chosen, and being limited by the Carnot efficiency.
You need to specify a CYCLE and plot temperature vs pressures around the FULL cycle. The work done per cycle is the area contained within that plot. To measure just say the expansion phase as you are here is meaningless in terms of determining the overall efficiency.
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Mark,
I took your response as a "slam" against Brad as well.
Tinman suffers a mental block in that he claims he can peroform work from the internal energy of a gas volume without losing the commensurate energy from that gas volume. Tinman insists on his claim in contravention to two hundred years of thermodynamics and numerous patient explanations. Anyone can suffer a mental block.
I think you did not understand when he asked you if you could calculate how much extra air would need to be added to make his 110% system readings work. Those were his readings but he did not know the increase in quantity.
I tire of tinman's habit of leaving out detail for whatever reason when he asks for opinions or calculations. It is a waste of everyone's time. I admire that he does experiments. For experiments to be meaningful they must be fully and openly described along with the data obtained and the control experiments run.
We all know that if more air is added into the system it changes everything, and to add that extra air takes energy, how much air was added?
Yes, tinman described one apparatus and an experiment and then offered data from a quite different apparatus. Then tinman objected when evaluation of the data he offered stated that the data was unreasonable for the apparatus that he had described. Well, duh, the data did not come from the apparatus he described and so there is no surprise that it was spotted as unreasonable based on what he described.
So,, he is setting the benchmark test runs,, you did the math for that, now he has made a change to the plumbing but not the vessels so volume stays the same and the first charged vessel still has the same quantity and pressure to start, the added quantity goes into the vessel that is being charged up by the charged vessel while it is being charged up.
If anyone wants ot conduct experiments and have those experiments reviewed then they need to describe their experiments accurately. You make an assumption and perhaps tinman did as well that the difference in apparatus did not matter. Well, obviously it did matter.
This added quantity of air is supplied by the outside environment and not the operator, hence no cost to the operator.
That's one point of view and there are a number of circumstances under which it would be true in practice. However, it is also true that the ability of the system to perform useful work decreased as a result of the operation, outside air admission notwithstanding.
Of course this extra quantity of air, under pressure, will need to be vented back to the outside environment to return the system back to the start conditions.
We have a starting condition of air at 40psi gauge in a 10l tank. Later we have 16.2psi gauge in 30l. PV product looks bigger doesn't it? So, can we now do the same thing again and end up with 50l at 11 or 12psi? Then 70 at say 9 or 10 psi gauge? Now for the really big question: Can we then take our higher volume, lower pressure gas and use it to reconstitute the 40psi gauge in the same or another 10l tank with anything left over? If we can then we have found a way to harvest heat energy from a single temperature reservoir of the outside air, have broken the second law of thermodynamics on a macro scale and can theoretically devise engines that will never need fuel and will combat global warming. Wouldn't that all be great stuff? There's just that one little problem: Entropy.
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The device is set up to do exactly was i have claimed it would do from the very start,and that is- draw in extra energy from the open part of the system(the enviroment) while increasing in energy.
I gave you the numbers above,and also the work that was done. The only reason that you do not wish to crunch those numbers is because the results will go against what you believe in.
The setup is as it was first described to be,with the addition of tank B.Tank A supplying a ram with compressed air. The ram doing usful work(lifting a 7kg weight 400mm),and the(now heated)gas returned to tank B. The rest of the opperation will be remaining with me pending your number crunching-If you will please :). Once i have the system to 150% by the numbers i supply you,i will then go about finding a way to present the system here on this thread so as it cannot be refuted. This will be done in a way that is to satisfy !your! guidlines Mark-->which i have tried to follow to the leter to this date.
You described one thing then supplied data from something else. When I noted that the data you supplied did not fit the thing that you described you objected that I am somehow stuck in a belief. The data will always tell the story. The data confirmed that the data you offered was inconsistent with the apparatus that you described.
The apparatus has yet to perform any work. When you get to the point that you get your apparatus to do work and measure the energy that it takes to get your apparatus through a complete cycle, then I will be happy to evaluate your data. What I am not interested in is where you describe one thing and supply data from something else.
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initially you did some work to raise the pressure in the tank, but conveniently ignore this in your specification of the 'system').
Absolute rubbish.
The work done to raise the pressure in tank A IS the stored energy in tank A. The efficiency of the compressor to perform this work has nothing at all to do with the efficiency of the DUT.
The best you will be able to achieve is that you can extract energy from the system equal to the amount of work performed to pressurise it initially,
Wrong.
The numbers Mark crunched on the DUT fitted with the venturi show's a 10% increase of energy above that of the starting value. This stored energy can now perform more usful work than that of the energy amount we started with.
Of course you can draw in extra energy from the environment by using some or all of the stored potential the cylinder at 40 psi. Then what?
Are you kiding me :o-->What next :o
Here you say it is possable to draw in extra energy from the enviroment,and then you ask !!Then what!!
Then we have more energy than what we started with. With this extra energy now avaliable,do you not think it possable that tank A can now be restored to it's original starting pressure,and we still have some stored energy left over in tank B?-->or dose this extra energy just up and vanish?
To be useful you have to be able to recompress that cylinder without using any FURTHER external work
See above statement.
You need to specify a CYCLE and plot temperature vs pressures around the FULL cycle. The work done per cycle is the area contained within that plot. To measure just say the expansion phase as you are here is meaningless in terms of determining the overall efficiency.
It's not meaningless at all. There is no point in going further until we have an increase in energy. This is like asking how fast the car go's before you know if the engine run's.
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Tinman suffers a mental block in that he claims he can peroform work from the internal energy of a gas volume without losing the commensurate energy from that gas volume. Tinman insists on his claim in contravention to two hundred years of thermodynamics and numerous patient explanations. Anyone can suffer a mental block. I tire of tinman's habit of leaving out detail for whatever reason when he asks for opinions or calculations. It is a waste of everyone's time. I admire that he does experiments. For experiments to be meaningful they must be fully and openly described along with the data obtained and the control experiments run.Yes, tinman described one apparatus and an experiment and then offered data from a quite different apparatus. Then tinman objected when evaluation of the data he offered stated that the data was unreasonable for the apparatus that he had described. Well, duh, the data did not come from the apparatus he described and so there is no surprise that it was spotted as unreasonable based on what he described.If anyone wants ot conduct experiments and have those experiments reviewed then they need to describe their experiments accurately. You make an assumption and perhaps tinman did as well that the difference in apparatus did not matter. Well, obviously it did matter.That's one point of view and there are a number of circumstances under which it would be true in practice. However, it is also true that the ability of the system to perform useful work decreased as a result of the operation, outside air admission notwithstanding.We have a starting condition of air at 40psi gauge in a 10l tank. Later we have 16.2psi gauge in 30l. PV product looks bigger doesn't it? So, can we now do the same thing again and end up with 50l at 11 or 12psi? Then 70 at say 9 or 10 psi gauge? Now for the really big question: If we can then we have found a way to harvest heat energy from a single temperature reservoir of the outside air, have broken the second law of thermodynamics on a macro scale and can theoretically devise engines that will never need fuel and will combat global warming. Wouldn't that all be great stuff? There's just that one little problem: Entropy.
TinMan disnt suffer any mental block !thank you!
I posted a sketch of the changes to the device. The sole purpose is to draw in energy from the enviroment(the open system),and increase the overall energy within the system.
The mental block is not on my side Mark,as you seem to be missing what was just achieved by adding the venturi to the system. The venturi effect was enabled by the gas flow from tank A. Work was being done due to the fact that the gas from the enviroment was being accelerated from a static point,through the venturi itself. Gas has mass,and work is done to accelerate a mass. The result was we had a situation where X amount of energy was supplied,and we ended up with Y amount of energy-a higher value of stored energy within the two tanks.
Can we then take our higher volume, lower pressure gas and use it to reconstitute the 40psi gauge in the same or another 10l tank with anything left over?
I believe yes,and this is why i asked you to crunch those last numbers for me-->so as i can show how tank A can be lifted back up to the starting pressure with tank B still having some pressurised gas left in it. Remember that tank A dose not have to be filled from scratch,as it still has 18psi of gas within it. Tank B which is twice the volume of tank A also hase 18psi of gas stored within it.
All i needed was those numbers so as i could calculate the needed rams to recompress tank A back to 40 psi from 18 psi. We have gained energy within the system,so there is no reason at all that tank A cannot be lifted back to it's starting pressure. As we now have a lower pressure to opperate any rams,we need to increase the size of the rams efective area to achieve the same force that can be supplied from the ram.
I am sure you want me to post the whole device and it's opperations-->but this just isnt going to happen until we have the completed device. So i ask that you work with the numbers i supply to you ,and that will be as has been-->the temperatures and pressures of each tank. The only energy entering the system will be enviromental energy,and the whole purpose of this project is to turn enviromental energy into usable energy. I know we can do this already with solar pannels,wind generators,and the like's-->but show me one that can do it 24 hours a day regardless of temperature,day/night or wind. We do have one of these already,and that is tidal power,but there seems to be few of these power generating systems being used.
The DUT is now as it was describe to be,consisting of the two tanks and a ram. There are ofcourse hoses,valves and fittings in the setup,but that is it. So as i claimed,i will use that ram to do usful work,and raise the energy level in the two tanks at the same time. I think you would agree that compressing gas is work being done.
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@Tinman
Nice work, and thank you for it.
I of course, hope that it continues to pan out well, and have every confidence that
as allways, you will to let the data tell the story.
@Others
Chill out, Tinman HAS NOT claimined OU. but rather he is exploreing a NOVEL approch to
tapping into a potential energy source. WTF
best wishes
floor
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HAS NOT claimined OU. but rather he is exploreing a NOVEL approch to
tapping into a potential energy source. WTF
best wishes
floor
I'm sorry but I don't see anything 'novel' here. It is not doing anything that has not been well understood the science of thermodynamics for two centuries.
If this system could cycle as he believes it would, then that would be a clear second law violation, and that by definition is 'OU'.
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Are you kiding me :o-->What next :o
Here you say it is possable to draw in extra energy from the enviroment,and then you ask !!Then what!!
Then we have more energy than what we started with. With this extra energy now avaliable,do you not think it possable that tank A can now be restored to it's original starting pressure,and we still have some stored energy left over in tank B?-->or dose this extra energy just up and vanish?
Yup that is exactly what I am saying. No matter what you do you will not be able to restore tank A to it's original starting position.
The only situation where you could approach that situation would be if the gas was expanded in a reversible adiabatic manner.
However, IF you did that you could also NOT draw any extra energy into the system. I know you don't believe it, but such a statement has two hundred years of thermodynamic theory and experimentation to back it up.
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Absolute rubbish.
The work done to raise the pressure in tank A IS the stored energy in tank A. The efficiency of the compressor to perform this work has nothing at all to do with the efficiency of the DUT.
Wrong.
The numbers Mark crunched on the DUT fitted with the venturi show's a 10% increase of energy above that of the starting value. This stored energy can now perform more usful work than that of the energy amount we started with.
Well the big problem here is entropy. You have managed to get the external system to add energy to your apparatus. The problem is how much useful work you can do, and how much work you will have to do in order to get back to your starting condition.
Are you kiding me :o-->What next :o
Here you say it is possable to draw in extra energy from the enviroment,and then you ask !!Then what!!
Then we have more energy than what we started with. With this extra energy now avaliable,do you not think it possable that tank A can now be restored to it's original starting pressure,and we still have some stored energy left over in tank B?-->or dose this extra energy just up and vanish?
Well since you think you can, then all you need to do now is to try and figure a way to do that.
See above statement.
It's not meaningless at all. There is no point in going further until we have an increase in energy. This is like asking how fast the car go's before you know if the engine run's.
Absolute energy is not the only issue at hand. Entropy is a big deal.
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Well the big problem here is entropy. . .Well since you think you can, then all you need to do now is to try and figure a way to do that.Absolute energy is not the only issue at hand. Entropy is a big deal.
Well the big problem here is entropy.
And why is entropy such a big issue? ,when the heat is both transferd into(entropy increases) and then disipated out of(entropy decreases) the system. The end temperature will always be the ambiant temperature in which the unit opperates.
The problem is how much useful work you can do, and how much work you will have to do in order to get back to your starting condition
If you would crunch those last numbers for me,i will have a starting point to work with. Im not asking you to verify the workings of some wonder machine(that bit will be up to me to do),i am asking for nothing more than the outcome or product of some results i have given.
.Well since you think you can, then all you need to do now is to try and figure a way to do that.
The way is figured out,but no more money will be spent until i know what i have in way of added energy. Im happy to spend the time and money,but i ask for help with the energy calculation's.
You have managed to get the external system to add energy to your apparatus
Yes,i have. But i need to know how much energy was added to the system before i can go any further. If you do not wish to do these calculations for me,then im sure there are others that will that reside in a different place. I have managed to amplify a given amount of energy at no cost to the original energy that created it. We now have a battery that self charges,a system that take a certain amount of energy,and increases it without loss. How can this not do more work than the original amount of energy we started with?. I can !without a doubt! show you the venturi effect increase a systems energy output while continually running,without any additional energy input-->would this be of any use?.The calculated input may be say 10 joules per second,while the output is say 15 joules per second. I will NOT however be held accountable for other devices inefficiencies-like a compressor.A compressor is not the DUT here,the same as a battery is not a battery charger.
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And why is entropy such a big issue? ,when the heat is both transferd into(entropy increases) and then disipated out of(entropy decreases) the system. The end temperature will always be the ambiant temperature in which the unit opperates.
You will find out soon enough when you try to put all those molecules back in Pandora's 10 liter cylinder.
If you would crunch those last numbers for me,i will have a starting point to work with. Im not asking you to verify the workings of some wonder machine(that bit will be up to me to do),i am asking for nothing more than the outcome or product of some results i have given.
Evaluating numbers is senseless if the apparatus involved is not specified. We've already been through that once. Describe the machine accurately and I will run numbers.
The way is figured out,but no more money will be spent until i know what i have in way of added energy. Im happy to spend the time and money,but i ask for help with the energy calculation's.
Yes,i have. But i need to know how much energy was added to the system before i can go any further. If you do not wish to do these calculations for me,then im sure there are others that will that reside in a different place. I have managed to amplify a given amount of energy at no cost to the original energy that created it. We now have a battery that self charges,a system that take a certain amount of energy,and increases it without loss. How can this not do more work than the original amount of energy we started with?. I can !without a doubt! show you the venturi effect increase a systems energy output while continually running,without any additional energy input-->would this be of any use?.The calculated input may be say 10 joules per second,while the output is say 15 joules per second. I will NOT however be held accountable for other devices inefficiencies-like a compressor.A compressor is not the DUT here,the same as a battery is not a battery charger.
Well what has happened is energy has been moved (as LE pointed out that's something that a heat pump does), and the entropy has been increased. There have been many people who thought that they could manipulate a heat pump: Device that expends useful work moving a quantity of heat energy between two reservoirs into a free energy machine. They run up against Sadi's Curse every time.
The percentage heat that can be obtained from a reservoir is the ratio of the temperature change incurred reducing the temperature by conveying work to the hot temperature. Because of that little formula PV = nRT, we know:
R is a constant, and if n doesn't change, then
PV = K1T. Consequently, as you increase the volume occupied by the gasses even with the same total energy, the usable energy that you can extract as work keeps shrinking. So somewhat analagous to a water ram where we can pump a small fraction of water above the starting level by dumping a whole bunch of water to a lower level, you have with the venturi and check valve pumped some of the surrounding atmospheric air up to a higher pressure at the cost of dumping a bunch of air from the 10 liter vessel to a much lower pressure.
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R is a constant, and if n doesn't change, then
PV = K1T.
You will find out soon enough when you try to put all those molecules back in Pandora's 10 liter cylinder.
There comes a time when the energy drawn in from the enviroment surpasses that lost to entropy.
There have been many people who thought that they could manipulate a heat pump:
Yes,i know. I have been involved in many different variations of heat pump devices,and my system is represents none of them.
Consequently, as you increase the volume occupied by the gasses even with the same total energy, the usable energy that you can extract as work keeps shrinking.
Yes,this we know. But as i stated above,there is a point where the energy gathered by the systems opperation surpasses the losses.
So somewhat analagous to a water ram where we can pump a small fraction of water above the starting level by dumping a whole bunch of water to a lower level, you have with the venturi and check valve pumped some of the surrounding atmospheric air up to a higher pressure at the cost of dumping a bunch of air from the 10 liter vessel to a much lower pressure.
Yes,the pressure is much lower,but the volume has increased by 3X. If we had some calculations done,we would then know if the now lower pressure but increased volume can do more work than that of the higher pressure but lower volume can do. We now have a 30 ltr tank with a pressure of 18psi+,when at the start we had a 10ltr tank with a pressure of 40.2psi. Which configuration could supply a device that needs a constant pressure of say 2psi at a fixed flow rate of say 1ltr a minute to opperate(do work) for a longer period of time?.Is it 10ltr's at 40psi,or 30ltr's at 18psi?.
Evaluating numbers is senseless if the apparatus involved is not specified. We've already been through that once. Describe the machine accurately and I will run numbers.
well it then seems we are done here,as like you said,we have been through this before,and i thought i made it very clear that the device and it's workings would not be disclosed until such time as i am ready to do so. I ask nothing more that for you to calculate start and end energies in both tank's,and for some reason you need to know exactly how the device is set up.
I will try once again-->we have our two tank's,and gas is transfered from one tank to another via way of the ram that is doing work as this transfer takes place. The ram starts and finishes each cycle empty,and as in the venturi setup,the ram is also drawing gas in from the enviroment,and pumping it into tank B. The venturi is only capable of raising the pressure in tank B so much,and then it can no longer contribute to the system. The ram however has the ability to lift the pressure to a much higher level in tank B before it will no longer opperate. Once there is not enough pressure in tank A to opperate the ram,then the valve between tank A and B is opened so as equilisation can take place-->this is the end of the process,we now have 18.2psi in each tank,with a combined volume of 30ltr's.
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Absolute rubbish.
The work done to raise the pressure in tank A IS the stored energy in tank A. The efficiency of the compressor to perform this work has nothing at all to do with the efficiency of the DUT.
Actually it have everything to do with 'it' unless you are going to simply ignore the fact that to operate a full cycle you MUST restore all the potential in the tanks back to the starting condition.
The compression part of the cycle has to be extremely efficient since the only net energy you have available is that,that was able to enter the system from the environment during the expansion phase. Unfortunately CARNOT rules here. That amount of 'ingested energy' will never be sufficient to perform the work required to recompress the system.
If you are simply going to ignore that requirement, then it is not even worth continuing to analyse. There has never been a dispute that you could extract work out of some store of potential without bothering to replenish it.
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There comes a time when the energy drawn in from the enviroment surpasses that lost to entropy.
Energy isn't lost to entropy. Entropy saps the ability to convert energy to useful work.
Yes,i know. I have been involved in many different variations of heat pump devices,and my system is represents none of them.
If you understand heat pumps, then you should understand the similarities here.
Yes,this we know. But as i stated above,there is a point where the energy gathered by the systems opperation surpasses the losses.
The normal way that we do honest accounting is to perform an energy balance through one or more complete cycles. Since we have not done that for your scheme at best you simply do not know where you are going to end up when an entire cycle is accounted. As it stands, you have greatly reduced the apparatus' capacity to perform useful work but you haven't done any useful work yet. As a Gedanken consider what it will take to end up with 10 l of 40psi gauge air at room temperature and how you will get there from where you are. You have 10 l of 16.2psi gauge, so roughly 30.9 psi absolute. You need to increase that gas quantity to 40 psi gauge or 54.7 psi absolute. You can either use both the 20l and 10l tanks to compress atmosphere into a 10l tank, or only the 20l tank contents to do work that will compress atmospheric gas into the 10l tank.
You will need to apply pressure at 40psi gauge + at the end of your recompression. How man psi gauge do you plan to have at that point? For sake of argument, say it is 5.0. You will need an 8:1 diaphragm or piston to develop the 40psi gauge from 5psi gauge. You surely don't have 8:1 volume to spare, so scratch that. Try 15.0 psi gauge. Now you need 8:3. But you can only afford to use about 6% of your gas, and you are again cooked. You could try and come up with a clever way to build a variable gain compressor so that just as you run out of gas you get back to 40 psi guage and hope with that you can get enough gas back into tank A. It won't happen.
Yes,the pressure is much lower,but the volume has increased by 3X. If we had some calculations done,we would then know if the now lower pressure but increased volume can do more work than that of the higher pressure but lower volume can do. We now have a 30 ltr tank with a pressure of 18psi+,when at the start we had a 10ltr tank with a pressure of 40.2psi. Which configuration could supply a device that needs a constant pressure of say 2psi at a fixed flow rate of say 1ltr a minute to opperate(do work) for a longer period of time?.Is it 10ltr's at 40psi,or 30ltr's at 18psi?.
While you object that I contend you suffer a mental block, entropy really seems to have you stumped. A bathtub full of room temperature water is useless towards driving a heat engine that only has other room temperature heat sources / sinks to work against. That bathtub has plenty of energy in it.
well it then seems we are done here,as like you said,we have been through this before,and i thought i made it very clear that the device and it's workings would not be disclosed until such time as i am ready to do so.
I have no interest in playing a game of GIGO. Calculations made based on grossly incomplete or incorrect assumptions yield silly results as we have already just seen.
I ask nothing more that for you to calculate start and end energies in both tank's,and for some reason you need to know exactly how the device is set up.
I will try once again-->we have our two tank's,and gas is transfered from one tank to another via way of the ram that is doing work as this transfer takes place. The ram starts and finishes each cycle empty,and as in the venturi setup,the ram is also drawing gas in from the enviroment,and pumping it into tank B. The venturi is only capable of raising the pressure in tank B so much,and then it can no longer contribute to the system. The ram however has the ability to lift the pressure to a much higher level in tank B before it will no longer opperate. Once there is not enough pressure in tank A to opperate the ram,then the valve between tank A and B is opened so as equilisation can take place-->this is the end of the process,we now have 18.2psi in each tank,with a combined volume of 30ltr's.
See above. I am not interested in GIGO. You may not consider that information you omit is significant. You just demonstrated that such details can be very significant.
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You have 10 l of 16.2psi gauge, so roughly 30.9 psi absolute. You need to increase that gas quantity to 40 psi gauge or 54.7 psi absolute. You can either use both the 20l and 10l tanks to compress atmosphere into a 10l tank, or only the 20l tank contents to compress do work that will compress atmospheric gas into the 10l tank.While you object that I contend you suffer a mental block, entropy really seems to have you stumped. A bathtub full of room temperature water is useless towards driving a heat engine that only has other room temperature heat sources / sinks to work against. That bathtub has plenty of energy in it..See above. I am not interested in GIGO. You may not consider that information you omit is significant. You just demonstrated that such details can be very significant.
The normal way that we do honest accounting is to perform an energy balance through one or more complete cycles.
In order for me to be able to complete the unit to perform a complete number,i need to know where i stand(how much energy i have)after the first half of the cycle-the half that gives us the energy to replenish the 10l tank to it's starting capacity. Without these numbers i cannot calculate the size or dimentions of the cylinder needed to recompress the gas into tank A.
Since we have not done that for your scheme at best you simply do not know where you are going to end up when an entire cycle is accounted.
This is correct,we do not have an answer due to you insisting on needing to know how the entire system opperates. You dont need to know this,i am asking only that you calculate the start and end energy amount avaliable in the two tanks. Knowing how the system opperates has nothing to do with the end result's. Im asking for end result's-not an evaluation on my device.
As it stands, you have greatly reduced the apparatus' capacity to perform useful work but you haven't done any useful work yet. As a Gedanken consider what it will take to end up with 10 l of 40psi gauge air at room temperature and how you will get there from where you are.
That is what im trying to do-work out how and what i need to get there from where i am ATM. However,i dont know where i am,as you are not interested in calculating the results i gave you for the first half of the cycle.
I have no interest in playing a game of GIGO. Calculations made based on grossly incomplete or incorrect assumptions yield silly results as we have already just seen
There are no silly game's other than the one you play Mark. Needing to know how the system work's has nothing to do with calculating the energy within the two tank's from the numbers i gave you-thats all i ask. Im not asking you to verify my device,nor am i after your opinion on how my device work's-->just calculating the start energy in tanks A&B,and the end results i have obtained in tanks A&B-thats all you have to do. But i see that this is going to be a never ending game of-->you show me first. Well it looks like it's game over here,and i must say that i have seen better styles of trying to get a glimps of a device and how it work's than the style you just used here. Im sorry,but i dont make deals,nor do i give in to demand's.
Thanks for your time.
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In order for me to be able to complete the unit to perform a complete number,i need to know where i stand(how much energy i have)after the first half of the cycle-the half that gives us the energy to replenish the 10l tank to it's starting capacity. Without these numbers i cannot calculate the size or dimentions of the cylinder needed to recompress the gas into tank A.
You are asking for piecemeal calculations against a model that is not in view. If you want to go calculate basic quantities on your own try the EngineeringToolBox.
This is correct,we do not have an answer due to you insisting on needing to know how the entire system opperates. You dont need to know this,i am asking only that you calculate the start and end energy amount avaliable in the two tanks. Knowing how the system opperates has nothing to do with the end result's. Im asking for end result's-not an evaluation on my device.
Excuse me but this is your project not mine. I did not entice you into trying something by making a promise that I have since broken.
That is what im trying to do-work out how and what i need to get there from where i am ATM. However,i dont know where i am,as you are not interested in calculating the results i gave you for the first half of the cycle.
There are no silly game's other than the one you play Mark. Needing to know how the system work's has nothing to do with calculating the energy within the two tank's from the numbers i gave you-thats all i ask. Im not asking you to verify my device,nor am i after your opinion on how my device work's-->just calculating the start energy in tanks A&B,and the end results i have obtained in tanks A&B-thats all you have to do. But i see that this is going to be a never ending game of-->you show me first. Well it looks like it's game over here,and i must say that i have seen better styles of trying to get a glimps of a device and how it work's than the style you just used here. Im sorry,but i dont make deals,nor do i give in to demand's.
Thanks for your time.
I did not lead you into this forest. LE did not lead you into this forest. Both of us have explained to you what's further down the road waiting for you. You have very confidently expressed that something entirely different lies ahead. Enjoy the journey.
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For those that are interested,here i show the venturi effect doubling the output of a small DC generator. This shows the venturi adding energy to the system,and doing work.
https://www.youtube.com/watch?v=xsDXKN4Mosw
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The demonstraton shows that more power transfers to the fan with the valve opened rather than closed. It does not determine that the power in the outflow exceeds the power consumed generating the inflow.
If I connect a DC power source to a fixed resistor and that to a variable resistor I can readily show that a value of the variable resistor exists that maximizes the power transferred to that variable resistor. What I cannot do is show that the power exceeds the power drawn from the power source. We would also see that the less power that the load draws from the power source the greater the percentage of power through the variable load resistor is of the diminishing power drawn.
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The demonstraton shows that more power transfers to the fan with the valve opened rather than closed. It does not determine that the power in the outflow exceeds the power consumed generating the inflow.
If I connect a DC power source to a fixed resistor and that to a variable resistor I can readily show that a value of the variable resistor exists that maximizes the power transferred to that variable resistor. What I cannot do is show that the power exceeds the power drawn from the power source. We would also see that the less power that the load draws from the power source the greater the percentage of power through the variable load resistor is of the diminishing power drawn.
The demonstraton shows that more power transfers to the fan with the valve opened rather than closed. It does not determine that the power in the outflow exceeds the power consumed generating the inflow
It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.
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It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.
This demonstrates nothing other than a more efficient nozzle. Try the same experiment without a venturi but simply vary the nozzle shape and diameter. You'll find exactly the same effect occurring. A nozzle that can generate a higher velocity flow will produce more lift on the fan and hence more output to the generator.
Since you are not actually measuring the energy potential of the source you can't make any pronouncement at all that an energy gain has occurred.
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It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.
Tinman there is no measure of the energy required to generate the flow in the tube. So that is a dead stop to determining outflow energy over inflow energy right there. It shows only what you see, more power transferred to the blades in one condition rather than the other. We can make the blade spin faster by increasing the power in the tube and/or by improving the impedance match between the blade and the outflow from the tube. Surely you agree that if we were to for example change the angle of the blades that we could change the fan speed a lot without hardly affecting the flow through the tube at all. We can turn that right around by changing the geometry of the nozzle.
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Tinman there is no measure of the energy required to generate the flow in the tube. So that is a dead stop to determining outflow energy over inflow energy right there. It shows only what you see, more power transferred to the blades in one condition rather than the other. We can make the blade spin faster by increasing the power in the tube and/or by improving the impedance match between the blade and the outflow from the tube. Surely you agree that if we were to for example change the angle of the blades that we could change the fan speed a lot without hardly affecting the flow through the tube at all. We can turn that right around by changing the geometry of the nozzle.
We could indeed change the angle of the blades to make it more efficient,but once again,the venturi open would still produce more torque from the fan. We have a higher gas flow with the venturi open than we do with it closed. Maybe we show the difference where flow angle or geometry dosnt matter. Maybe we show usful work being done by way of the flow of gas lifting a mass in a given amount of time?. It takes more energy to accelerate a given mass in a shorter period of time-dose it not?. So how about we do something really simple,like put a pingpong ball(our mass),into a tube of a given length,and see how fast it travels through that tube with and without the venturi open?.
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We could indeed change the angle of the blades to make it more efficient,but once again,the venturi open would still produce more torque from the fan. We have a higher gas flow with the venturi open than we do with it closed. Maybe we show the difference where flow angle or geometry dosnt matter. Maybe we show usful work being done by way of the flow of gas lifting a mass in a given amount of time?. It takes more energy to accelerate a given mass in a shorter period of time-dose it not?. So how about we do something really simple,like put a pingpong ball(our mass),into a tube of a given length,and see how fast it travels through that tube with and without the venturi open?.
Changing a variable and noting that the fan speeds only tells us that more power was imparted to the fan blades. Since that can occur for several reasons, it does not tellwhich of those reasons or combinations of those reasons.
Isn't your hypothesis that you can perfom more work than energy you input by plumbing your apparatus to the local environment? If that is what you are trying to get to, then the experiment design effort should try and falsify that as directly and simply as possible.
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TinMan
Very cool Video
https://www.youtube.com/watch?v=xsDXKN4Mosw
of course some don't seem to appreciate your intuition and hard work, Personally
I know that you have many "fans" here that do appreciate your line of thought.
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TinMan
Very cool Video
https://www.youtube.com/watch?v=xsDXKN4Mosw
of course some don't seem to appreciate your intuition and hard work, Personally
I know that you have many "fans" here that do appreciate your line of thought.
Chet, the trick is to get at the truth.
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Isn't your hypothesis that you can perfom more work than energy you input by plumbing your apparatus to the local environment?
As clearly stated in the video,the video was to show effect,not efficiency of the system.
If that is what you are trying to get to, then the experiment design effort should try and falsify that as directly and simply as possible.
Try and falsify?. How would one do that when it can be seen as clear as day?. That would be like me trying to show that a 12 volt battery really dosnt have 12 volts across it's terminals.
Changing a variable and noting that the fan speeds only tells us that more power was imparted to the fan blades. Since that can occur for several reasons, it does not tellwhich of those reasons or combinations of those reasons.
This is why i suggested the mass acceleration test. Would you not agree that to accelerate a mass to a higher speed over a set distance requires more energy ?.
I can remove the nozzel from the venturi,and point that at the fan blade so as it's in the very same position it was when housed in the venturi,and see what the outcome is-->but of course,i have already done this,and can tell you the results are exactly as they are when the nozzel is housed in the venturi.
If you would like to put forth a test setup Mark,i would be more than happy to try it out.
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This demonstrates nothing other than a more efficient nozzle.
Since you are not actually measuring the energy potential of the source you can't make any pronouncement at all that an energy gain has occurred.
Try the same experiment without a venturi but simply vary the nozzle shape and diameter.You'll find exactly the same effect occurring. A nozzle that can generate a higher velocity flow will produce more lift on the fan and hence more output to the generator.
Sure LE.
We can also bore out an engine,put a set of high flow high compression heads on it,and say we now have more HP for nothing-->forgetting the fact that it now consumes more fuel to gain that extra HP.
We can reshape,change diameters of the nozzel to what ever shape you want,and then get the maximum efficiency. If we then take that same nozzel and place it in a venturi setup such as the one i showed,the fan would still spin faster.
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Sure LE.
We can also bore out an engine,put a set of high flow high compression heads on it,and say we now have more HP for nothing-->forgetting the fact that it now consumes more fuel to gain that extra HP.
We can reshape,change diameters of the nozzel to what ever shape you want,and then get the maximum efficiency. If we then take that same nozzel and place it in a venturi setup such as the one i showed,the fan would still spin faster.
The problem here is you are not measuring the 'fuel consumption'. Consider the air hose is supplied by a compressed tank that and the air exits out various shaped nozzles. Obviously the time take to empty the tank would be dependent on the size and design of the nozzle. The amount of 'fuel' would be constant but the total energy generated by would be dependent on runtime and the efficiency of the conversion performed by fan.
Measuring instantaneous power is not sufficient to claim an ENERGY increase.
Also, once you have exhausted the tank you then need to recompress the tank. Did you create enough stored potential to do that.? I can tell you for certain that you never can, no matter how efficient your nozzle is or whether it contains a venturi or not.
None of this of course precludes you from finding and environmentally supplied source of heat and a corresponding environmental sink at a lower temperature and devising a heat engine that operates between those two reservoirs.
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As clearly stated in the video,the video was to show effect,not efficiency of the system.
I am trying to avoid spending a bunch of time on tests that will be difficult to perform properly just to get an intermediate result. As I explained, if you really want to know what is going on with this set-up beyond the obvious A-B comparison that it imparts more power to the fan, then you've got to measure several variables for each of the three ports which would be a PITA.
Try and falsify?. How would one do that when it can be seen as clear as day?. That would be like me trying to show that a 12 volt battery really dosnt have 12 volts across it's terminals.
All good experiments are designe to falsify a hypothesis. That's how the scientific method works: Come up with a hypothesis. Design experiments that can disprove the hypothesis. Conduct those experiments. And if none of the experiments falsify the hypothesis then the hypothesis is accepted as true. It is really, really important to understand this process and follow it.
This is why i suggested the mass acceleration test. Would you not agree that to accelerate a mass to a higher speed over a set distance requires more energy ?.
I can remove the nozzel from the venturi,and point that at the fan blade so as it's in the very same position it was when housed in the venturi,and see what the outcome is-->but of course,i have already done this,and can tell you the results are exactly as they are when the nozzel is housed in the venturi.
At a minimum you want to measure the energy into the tube and energy out of the tube to enough accuracy that you can distinguish between the results with and without the local air feed. The values are kind of small making this a PITA on a budget.
If you would like to put forth a test setup Mark,i would be more than happy to try it out.
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The problem here is you are not measuring the 'fuel consumption'. Consider the air hose is supplied by a compressed tank that and the air exits out various shaped nozzles. Obviously the time take to empty the tank would be dependent on the size and design of the nozzle. The amount of 'fuel' would be constant but the total energy generated by would be dependent on runtime and the efficiency of the conversion performed by fan.
Measuring instantaneous power is not sufficient to claim an ENERGY increase.
Also, once you have exhausted the tank you then need to recompress the tank. Did you create enough stored potential to do that.? I can tell you for certain that you never can, no matter how efficient your nozzle is or whether it contains a venturi or not.
None of this of course precludes you from finding and environmentally supplied source of heat and a corresponding environmental sink at a lower temperature and devising a heat engine that operates between those two reservoirs.
The thing is LE,we could make what ever shape or size nozzel we like,but as soon as the gas leaves that nozzel it will draw in ambiant gas(air) with it. The venturi device isnt what makes the venturi effect,the venturi device was based around an effect that already exist. The reason for the venturi device is to localise that effect into a small tube so as it can be put to use. In order to see what only the compressed gas can do as far as work go's ,then it must be isolated from the enviroment when leaving the nozzel,so as it cant draw in the ambiant gas that surrounds it.
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I am trying to avoid spending a bunch of time on tests that will be difficult to perform properly just to get an intermediate result. As I explained, if you really want to know what is going on with this set-up beyond the obvious A-B comparison that it imparts more power to the fan, then you've got to measure several variables for each of the three ports which would be a PITA.All good experiments are designe to falsify a hypothesis. That's how the scientific method works: Come up with a hypothesis. Design experiments that can disprove the hypothesis. Conduct those experiments. And if none of the experiments falsify the hypothesis then the hypothesis is accepted as true. It is really, really important to understand this process and follow it.At a minimum you want to measure the energy into the tube and energy out of the tube to enough accuracy that you can distinguish between the results with and without the local air feed. The values are kind of small making this a PITA on a budget.
So we have X amount of kinetic energy from the nozzel,and we have that pesky law-->the conservation of energy. But the fan blades rely on a given amount of force applied to them in order to be able to do usful work. Now,there is no conservation of force law,and that is where we may have a win situation. So it takes X amount of applied force to lift say a 1kg ball 1 meter vertically into the air-which is 9.8 joules of energy. But if we roll that ball up a 45* incline to a hight of 1 meter,we now have to apply less force to raise that ball 1 meter,and the energy required to do so remains the same. With the fan,the blades are that incline,BUT that incline is now moving in the opposite direction to the applied force.
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The thing is LE,we could make what ever shape or size nozzel we like,but as soon as the gas leaves that nozzel it will draw in ambiant gas(air) with it. The venturi device isnt what makes the venturi effect,the venturi device was based around an effect that already exist. The reason for the venturi device is to localise that effect into a small tube so as it can be put to use. In order to see what only the compressed gas can do as far as work go's ,then it must be isolated from the enviroment when leaving the nozzel,so as it cant draw in the ambiant gas that surrounds it.
Ok, for interest sake lets assume you are correct and there there is an increase of energy occurring across the device. Surely then we should be able to chain these devices together and record an energy increase at each one?
If that occurs then why not chain them in circle ? Then just a small amount of input energy would be quickly transformed into a roaring vortex as one device amplified the output of the previous one.
I think you know intuitively that would not occur. If you believe it could then it is probably the simplest way to test your hypothesis. No complex measurements and calculations of temperature and pressure required. No every increasing vortex = no energy increase. Self accelerating vortex = energy increase.
If you can show the latter I'd happily admit 200 years of thermodynamics are wrong.
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So we have X amount of kinetic energy from the nozzel,and we have that pesky law-->the conservation of energy. But the fan blades rely on a given amount of force applied to them in order to be able to do usful work. Now,there is no conservation of force law,and that is where we may have a win situation. So it takes X amount of applied force to lift say a 1kg ball 1 meter vertically into the air-which is 9.8 joules of energy. But if we roll that ball up a 45* incline to a hight of 1 meter,we now have to apply less force to raise that ball 1 meter,and the energy required to do so remains the same. With the fan,the blades are that incline,BUT that incline is now moving in the opposite direction to the applied force.
Tinman there are many machines that take advantage of force manipulations: fan jets, the Dyson "Air Multiplier", etc. None of those machines generate surplus energy. They limit the losses for the tasks they do.
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It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done.
By the fan, which means more energy is being transferred to the fan in one condition than another. We cannot however conclude from that information that there is any energy (steady power) surplus.
The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed.
Your accounting is wrong. Take your handy compressor. Attach a 1m hose. Measure the outflow and pressure. Attach a 10m hose. Repeat the measurements. Energy is still conserved. Account for the observations.The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.
You can perform comparative measurements all day long and they are not going to tell you what you want to know.