PopularFX
Home Help Search
Advanced search 
Login Register
Welcome,Guest. Please login or register.
2025-03-17, 05:40:13
News: A feature is available which provides a place all members can chat, either publicly or privately.
There is also a "Shout" feature on each page. Only available to members.

Pages: [1] 2
Author Topic: Non Conservation of Energy Research  (Read 6482 times)

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
A place to put forward devices that seem to defy that law--the conservation of energy.

To add a device here that fits the above description,it must be replicable.
All comments and mathematical evaluations welcome,as long as they are accurate,and agree with data collected from actual devices.

No imaginary equalization math will be accepted here-->the data collected from actual devices will be the priority source of calculated results.

In the next few posts,i will be presenting 3 different test beds for tank to tank transfer of compressed gases.
In all 3 we will see either an increase in stored  energies,or energy dissipated without loss of stored energy.

These are actual devices i have built,and are easily replicable for 3rd party verification.
With these tank to tank transfer experiments,we follow the below

1- Temperatures will remain a constant,where at the start of each test, the compressed gas temperature in tank A will be allowed to settle to ambient temperature(room temperature)
At the end of each test,the gas temperature in both tanks will once again be allowed to settle to ambient temperature.
This is to eliminate the calculation of dissipated heat energies during each test.

2- The pressure values will be kept low,so as our compressed air remains withing the ideal gas region.
Air will retain a compressibility value of 1(or very close to) when the pressure is under 1400psi-->we will be well below this value.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
Our first DUT is as follows

We have two 50 liter tanks.
Between the tanks,we have a transfer pipe with a valve inline.
While the valve is closed,tank A is pressurized to 40psi gauge pressure at ambiant temperature.
Tank B is at 0 gauge pressure.
The valve is then open,and air begins to flow from tank A to tank B
The flow will stop when both tanks have an equal pressure of 20 psi gauge pressure,and are equal in temperature-ambient temperature.
The now PE contained within the two tanks is equal to that of what the PE of tank A was before the valve was opened-->energy was conserved.
See first diagram.

The second test is the same as the first,but where we now have an air motor between the transfer valve and tank B.
Coupled to this air motor is a small generator,which has a light bulb as a load across it's output.
The valve is opened,and air starts to flow from tank A to tank B.
But this time we also have the air motor spinning,driving the generator thats powering the lightbulb.

The whole time the air is flowing from tank A to tank B,energy is being dissipated from the lightbulb.
Regardless of this dissipated energy,the air will once again stop flowing when both tanks reach a gauge pressure of 20psi,and ambient temperature.
Energy was not conserved
See second diagram.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Interesting experiment Brad. I once upon a time had the idea for a bicycle that would operate on this principle of air shuttling between tanks via an air motor, and also allow regenerative braking, storing braking energy as pressure in the drive tank and vacuum tank, also storing excess pedaling energy on the level road to be used later for the hill climb. I mused that aluminum storage tanks would be much lighter than lead acid batteries. Also since the same dry air is shuttled corrosion effects would be minimized.

I gave up on the idea as I felt it would be too lossey due to the heat of compression. Now it occurs to me that you have an additional untapped energy source that can be captured with a (separated)differential thermopile, because as the gas in tank A momentarily cools due to expansion, the gas in tank B is momentarily heated due to compression. This is tapped with basically what is a differential thermocouple circuit, but can also be done locally with a thermopile known as a peltier module.

Quote
Thermopiles do not respond to absolute temperature, but generate an output voltage proportional to a local temperature difference or temperature gradient.

The EMF can also be obtained by extending the thermocouple circuit's differential junctions for non-local extraction. Thermopiles are not that efficient, so you might be better off using a Stirling Engine that can be arranged to do double duty as an air motor and heat differential engine. Such an idea might be patent worthy. Granted the temperature differential is momentary but every little bit helps.

It is good to note that the compression or expansion is actually the release or absorption of electromagnetic energy in the far infrared region.
This effect is in itself interesting to contemplate, e.g. why it occurs, and why it does not persist under constant pressure? Why does it equalize? What is the conversion process from mechanical pressure to electromagnetic energy? What is being given up or taken in? Why does it occur in all objects, even hard steel bars subject to pressure or the reverse?

To get a little more precise with your suggested test, may I suggest the use of electronic temperature compensated pressure sensors, as the ordinary gauges used on air tanks are not that accurate, precise, or repeatable due to friction and stiction in the gear drive (why you need to tap on them to get a better reading), and also subject to observer parallax errors. You can get these electronic sensors as stand alone analog or digital types to interface with Arduino, which will also provide a graphing and logging function for you.

May I also suggest that each of these non-conservation attempts be given a separate sub-thread so there is no cross-confusion when discussing them. I am very interested in the subject presented.

regards
« Last Edit: 2018-06-24, 19:45:54 by ion »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Group: Experimentalist
Hero Member
*****

Posts: 1858
Brad,

No math used but I would like to apply some logic if I may.  In all my work thru the years on cars, I have had lots of opportunity to experience the labor of my air compressor as the pressure builds in the tank and I'm sure you have too.  This tells me that it will take more energy to produce 40 psi in a given cylinder than 20 psi and if this analogy is equivalent to charged capacitors, then with both cylinders equalized to 20 psi, you will have half the total energy left compared to the start and the apparent lost energy is what's powering the air motor/gen.  Jus saying. ???

Regards,
Pm
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
Brad,

No math used but I would like to apply some logic if I may.  In all my work thru the years on cars, I have had lots of opportunity to experience the labor of my air compressor as the pressure builds in the tank and I'm sure you have too. This tells me that it will take more energy to produce 40 psi in a given cylinder than 20 psi and if this analogy is equivalent to charged capacitors, then with both cylinders equalized to 20 psi, you will have half the total energy left compared to the start and the apparent lost energy is what's powering the air motor/gen.  Jus saying. ???

Regards,
Pm

If this is true,then why is the same result had in the same test that has no air motor?--where did that energy go?

Something else to think about-->we charge the air tanks,and in the cap to cap transfer,no !charge! is lost.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Brad,

No math used but I would like to apply some logic if I may.  In all my work thru the years on cars, I have had lots of opportunity to experience the labor of my air compressor as the pressure builds in the tank and I'm sure you have too.  This tells me that it will take more energy to produce 40 psi in a given cylinder than 20 psi and if this analogy is equivalent to charged capacitors, then with both cylinders equalized to 20 psi, you will have half the total energy left compared to the start and the apparent lost energy is what's powering the air motor/gen.  Jus saying. ???

Regards,
Pm

Agreed, we all remember all the hoopla surrounding discharging capacitors, one into the other, some using a motor to do so.

Are the charging curves similar for air vs capacitors? Can we build a very close model to air cylinders using capacitors and other passive components?

Maybe Brad has something to add to the experiment using air and cylinders. Good data should tell the story.

I'm also very interested in any answers to some of the questions I posed in my first reply.

regards

p.s. I was writing while Brad was posting, sorry.
« Last Edit: 2018-06-25, 13:29:47 by ion »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
For those that think the tank to tank transfer of compressed gas may be the same as the cap to cap transfer,where you loose half the energy during the transfer,then i present the below.

This will be my second example where i show potential energy producing dissipated energy,without loss of that potential energy.

In this test,we use water instead of a compressed gas,and buckets instead of sealed tanks.

Two buckets sit on a table 1 meter above the ground
Each bucket weighs 1 KG.
A pipe with inline valve links the two buckets.
Test 1
Bucket A starts of with 20 liters of fresh water,and bucket B is empty.
The PE value is 215.82 joules
The valve is opened,and water flows from bucket A into bucket B.
The flow stops when both buckets have leveled out at 10 liters each.
The PE value is still 215.82 joules.
Energy was conserved.

Test 2
Test 2 is the same as test 1,but where a water turbine and generator are added inline between the valve and bucket B. A lightbulb is placed across the generator output as a load.
As before,bucket A has 20 liters of water in it,and bucket B is empty.
The PE value is 215.82 joules.
The valve is opened,and water flows through the turbine,and into bucket B.
The turbine turns the generator,which in turn lights the bulb.
The flow stops when both buckets have leveled out at 10 liters each.
The PE value is still 215.82 joules.
Energy was not conserved


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
 author=ion link=topic=3640.msg68502#msg68502 date=1529847442]

Quote
Interesting experiment Brad. I once upon a time had the idea for a bicycle that would operate on this principle of air shuttling between tanks via an air motor, and also allow regenerative braking, storing braking energy as pressure in the drive tank and vacuum tank, also storing excess pedaling energy on the level road to be used later for the hill climb. I mused that aluminum storage tanks would be much lighter than lead acid batteries. Also since the same dry air is shuttled corrosion effects would be minimized.

That is a good idea,and tanks lighter and stronger than ali are available today--carbon fibre is the way to go  O0
Heat loss can be kept to a minimum with slow compressing of the gas. So what you would want,is a small compressor coupled to your chain drive system,and a large tank--even one that go's in a small trailer behind the bike. With some smart engineering,the compressor could also be your motor.

Quote
Now it occurs to me that you have an additional untapped energy source that can be captured with a (separated)differential thermopile, because as the gas in tank A momentarily cools due to expansion, the gas in tank B is momentarily heated due to compression. This is tapped with basically what is a differential thermocouple circuit, but can also be done locally with a thermopile known as a peltier module.

Indeed
The tanks could be made from RHS,and the peltier modules sandwiched in between the two tanks.
This would give you an AC output from the peltier modules when you hit the transfer valve.
One tank starts of hot,and the other cold. As one id decompressing and cooling,the other is compressing and heating.
That is one way of converting some of the heat losses into usable power.

Quote
It is good to note that the compression or expansion is actually the release or absorption of electromagnetic energy in the far infrared region.
This effect is in itself interesting to contemplate, e.g. why it occurs, and why it does not persist under constant pressure? Why does it equalize? What is the conversion process from mechanical pressure to electromagnetic energy? What is being given up or taken in? Why does it occur in all objects, even hard steel bars subject to pressure or the reverse?

Above my pay grade ATM  :D

Quote
To get a little more precise with your suggested test, may I suggest the use of electronic temperature compensated pressure sensors, as the ordinary gauges used on air tanks are not that accurate, precise, or repeatable due to friction and stiction in the gear drive (why you need to tap on them to get a better reading), and also subject to observer parallax errors. You can get these electronic sensors as stand alone analog or digital types to interface with Arduino, which will also provide a graphing and logging function for you.

Ahhhh,im a mechanical man.
All that fancy digital-electronic mumbo jumbo is just more work ontop of work i already have in front of me--not to mention,more cash outlay.

I have a couple of high grade-low pressure-large in size gauges of an oxy set.
These ones are quite accurate,and would do the job well enough in these experiments  O0

Quote
May I also suggest that each of these non-conservation attempts be given a separate sub-thread so there is no cross-confusion when discussing them. I am very interested in the subject presented.

Well,i will keep all the air tank ones together here,and maybe a new thread on the water tank ones--but there pretty simple,and easy to see that there is no energy loss during each transfer,whether the turbine and generator are included or not.

In fact,if what Smudge says about the higher center of mass having a lower PE value,then when the water drains from one bucket to the other,we are lowering the center of mass,and so increasing it's potential energy value.  O0

Now,about that energy loss in the cap to cap transfer  ^-^

Is there a thread on this somewhere we can move this subject to?-->better brace yourself,as i have a theory on that,and some will not be happy  C.C


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Experimentalist
Hero Member
*****

Posts: 1858
Brad,

There is a somewhat interesting thread that has started on EF regarding water transfer here-

http://www.energeticforum.com/renewable-energy/20152-tanked-water-free-electricity-complete-home-generation.html#post311384

The gentleman claims OU and is supposedly running his farm with the device!

Regards,
Pm
   

Group: Professor
Hero Member
*****

Posts: 1953
This site shows a machine doing what is suggested on this thread http://www.ftexploring.com/energy/2nd_Law.html
in language that is easy to understand.  It seems you can do what Brad suggests, but you can't in practice get back to the starting conditions, you can do it only once.  If there were no losses you could use the gained energy to pump it back to starting conditions, but that is not OU, you haven't gained any useful energy.  You could go on from Brads two tanks at half pressure and discharge them into two more tanks, and keep doing this until you had almost zero pressure in a enormously large tanks.  But then what?  You can't get back to square one and do it all again.

Smudge
   
Group: Ambassador
Hero Member
*****

Posts: 4051
Partzman
I did reach out for that fellow when he first posted that claim.... or his experience.

From his contributions at different times I believe he is not too far away from where I live in NE USA ?

never heard back from him , nor did he ever respond to anyone on that thread .
« Last Edit: 2018-06-25, 18:58:47 by Chet K »
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Brad said:

Quote
For those that think the tank to tank transfer of compressed gas may be the same as the cap to cap transfer,where you loose half the energy during the transfer,then i present the below.

You lose a portion of the energy in a cap to cap transfer if you do not do it correctly e.g. a direct sparking wire connection. If you do it correctly with an inductor, energy is conserved. A low loss diode plus inductor is even better. You need some compliance in the circuit in the form of a large inductive reactance between the capacitors otherwise the very large current spike winds up radiating energy as radio waves and there is Joule heating in the wires and the weld spot of contact accounting for the losses.

The air tanks don't have the equivalent of a current spike problem as the space in the tank and compressible gas provide the required compliance. The air tank transfer has other losses that can possibly be partially recovered.

I implied there is some similarity of past attempts at the 2 caps + motor scenario to the air tanks, but agree they are not exactly the same.

regards
« Last Edit: 2018-06-26, 02:32:32 by ion »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   

Group: Tinkerer
Hero Member
*****

Posts: 3948
tExB=qr
potential energy is "relative"

The PE for 20 liters of water, 1 meter above the ground is gravitational potential energy relative to the surface earth.  This is not relative to the second tank.

   

Group: Mad Scientist
Full Member
***

Posts: 185
Reminds me of a similar iteration using only one tank and displacement. :)

https://overunity.com/6836/knitels-infinitypump/msg158166/#msg158166


---------------------------
When you say something is impossible, you have made it impossible
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
 author=ion link=topic=3640.msg68527#msg68527 date=1529959348]


Quote
You lose a portion of the energy in a cap to cap transfer if you do not do it correctly e.g. a direct sparking wire connection. If you do it correctly with an inductor, energy is conserved. A low loss diode plus inductor is even better. You need some compliance in the circuit in the form of a large inductive reactance between the capacitors otherwise the very large current spike winds up radiating energy as radio waves and there is Joule heating in the wires and the weld spot of contact accounting for the losses.





How about this--
Even if we use super conducting wire,with an ideal switch,we still find that half the energy was lost in the cap to cap transfer  ???
There is a reason for this,and it is not due to resistive losses.

There is also a reason that using inductive transfer methods yields better results in the energy transfer.

How about electromagnetic radiation losses O0
This would explain many things in regards to the cap to cap transfer losses.

1-If you use a low resistance transfer path,you increase the current flow,and also the amplitude of the electromagnetic field. Increasing the amplitude of the electromagnetic field requires more energy,and so more energy is dissipated by the electromagnetic field.

2-If you use a high resistance transfer path,you decrease the current,and so, decrease the electromagnetic radiation,but at the same time,you have increased the resistive losses.

When you hook two caps together,you have just made a loop,and current flowing through a loop creates an electromagnetic field-->and this is a DC current flow-->best kind of electromagnet for turning most of the energy into heat-->worst kind to use in the cap to cap transfer.

Which is the most efficient kind of electromagnet ?--> the AC kind of course.
DC causes a lot of heat,and the field is radiated away from the source.
But with AC,the field is pushed out,and then returns back to the source.
Very little is lost to heat and radiation.

Are we starting to see why using an inductive transfer in the cap to cap energy transfer is better?
Yes,the electromagnetic radiation produced by the circuit-and the internal plates of the capacitors,is returned back to it's source,as the inclusion of the inductor creates an AC wave.

Quote
The air tank transfer has other losses that can possibly be partially recovered.

Could you touch on those losses please ION
Heat has already been covered,and the only other loss i could think of is entropy.
But after much discussion with MarkE about the entropy issue,it was deemed not to be of any significant's.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
This site shows a machine doing what is suggested on this thread http://www.ftexploring.com/energy/2nd_Law.html
in language that is easy to understand.  It seems you can do what Brad suggests, but you can't in practice get back to the starting conditions, you can do it only once.  If there were no losses you could use the gained energy to pump it back to starting conditions, but that is not OU, you haven't gained any useful energy.  You could go on from Brads two tanks at half pressure and discharge them into two more tanks, and keep doing this until you had almost zero pressure in a enormously large tanks.  But then what?  You can't get back to square one and do it all again.

Smudge

Depends on how you look at it.

What if we could transfer half the stored energy from one cap to another,while driving a load that is in series between the two caps,and end up with the same amount of energy that we started with ?.
We still have to charge the first cap up to start with,just like we do with the air tank.

If some one was to do this with capacitors,you would be asking yourself--where did the energy come from that was driving that load during the transfer,when the stored energy at the end of the transfer was exactly the same as the stored energy at the start of the transfer.
So i would ask--why is it different when using air and tanks?.

As you have acknowledged,it dose work the way i say it dose.
Regardless of whether or not there is a load placed between the tanks transfer line,the end result is the same.
So you have to ask--where did the energy come from that was driving that load,if no energy was lost in the process  :o

You have to remember that the tanks are energy storage devices--just like a capacitor and battery.
And just like the capacitor and battery,we have to charge them up.

Quote
but you can't in practice get back to the starting conditions

I wouldnt be to sure about that  ;)


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Jr. Member
**

Posts: 66
Quite an interesting subject... I've experimented with various ideas along the same line of thoughts.   One of which I called the charge pump although it might be better named a charge shuttle.   The diagram below shows the flow of energy through one full switching cycle.

All the caps ( C1,C2,C3 ) are identical and charged to 36 volts - by starting with an equal amount of energy in the caps we eleviate a large portion of the inherent losses caused from discharging a full cap into an empty cap.
 
Most is self explanatory as it proceeds through the cycle.   What isn't quite as transparent is the fact that C2,C3 never loose their initial charge, they are never discharged - the energy is simply shuttled back and forth through the load.   If we use a watt meter on the input side ( C1 )  and a meter on the load we would see that the input is measured once per cycle but the load would be measuring both movements between charge and re-balance.

If we use a battery in place of C1 and  large capacitors for C2,C3 the benefits of shuttling energy increases.   The drawbacks are that the circuit acts as a current amplifier and voltage divider - generally the load sees 60% or less of the initial voltage but up to 2.6 times the current.   If low resistance loads are used the current can reach extremely high levels through the system, the charge/discharge cycles can become quite violent (massive currents) if not controlled.   

C1 or a battery in its place would be the "pressure" vessel in this scenario and does require recharging.   Although, I can show examples where input/output are closely matched. 
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
Quite an interesting subject... I've experimented with various ideas along the same line of thoughts.   One of which I called the charge pump although it might be better named a charge shuttle.   The diagram below shows the flow of energy through one full switching cycle.

All the caps ( C1,C2,C3 ) are identical and charged to 36 volts - by starting with an equal amount of energy in the caps we eleviate a large portion of the inherent losses caused from discharging a full cap into an empty cap.
 
Most is self explanatory as it proceeds through the cycle.   What isn't quite as transparent is the fact that C2,C3 never loose their initial charge, they are never discharged - the energy is simply shuttled back and forth through the load.   If we use a watt meter on the input side ( C1 )  and a meter on the load we would see that the input is measured once per cycle but the load would be measuring both movements between charge and re-balance.

If we use a battery in place of C1 and  large capacitors for C2,C3 the benefits of shuttling energy increases.   The drawbacks are that the circuit acts as a current amplifier and voltage divider - generally the load sees 60% or less of the initial voltage but up to 2.6 times the current.   If low resistance loads are used the current can reach extremely high levels through the system, the charge/discharge cycles can become quite violent (massive currents) if not controlled.   

C1 or a battery in its place would be the "pressure" vessel in this scenario and does require recharging.   Although, I can show examples where input/output are closely matched.

Interesting simple circuit.
Might give it a shot,but where C1 is swapped out for a 1.5v battery,and C2,C3 swapped out for two 50F caps.

Maybe even an inductor between C1 and SW2/C3  :D


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite
Hero Member
******

Posts: 3537
It's turtles all the way down
Brad, you said:

Quote
How about this--
Even if we use super conducting wire,with an ideal switch,we still find that half the energy was lost in the cap to cap transfer  ???
There is a reason for this,and it is not due to resistive losses.

There is also a reason that using inductive transfer methods yields better results in the energy transfer.

How about electromagnetic radiation losses O0
This would explain many things in regards to the cap to cap transfer losses.

I see a deeper understanding would be helpful. When I said radio waves, I implied electromagnetic radiation.

Even superconducting connecting wires would have a small inductance, and that small inductance would cause the wires to radiate at very high frequency, and the radiation efficiency would be high because at high frequencies you need only very short wires to be efficient antennas.

However if a large inductor (100's of mH) is used, the frequency will be very low and the short wires are very inefficient radiators (antennas)  at low frequencies, producing poor coupling to the ether. This assumes the inductor is a closed magnetic path. If it is an open magnetic path such as a ferrite rod, the radiation efficiency will go up.

Try to remember in a cap to cap transfer the current can theoretically approach infinity for a very brief instant so it excites the tank coil to radiate EM at a high value.

hope that helps
regards
« Last Edit: 2018-06-26, 17:50:31 by ion »


---------------------------
"Secrecy, secret societies and secret groups have always been repugnant to a free and open society"......John F Kennedy
   
Jr. Member
**

Posts: 66
Interesting simple circuit.
Might give it a shot,but where C1 is swapped out for a 1.5v battery,and C2,C3 swapped out for two 50F caps.

Maybe even an inductor between C1 and SW2/C3  :D


Brad

The 50F caps might require very slow switching to move energy from one to the other.   With 1F caps you can move energy fairly well at 5 to 15 hz  https://www.youtube.com/watch?v=XTVbTQnRmKM

Ed
   

Group: Professor
Hero Member
*****

Posts: 1953
So you have to ask--where did the energy come from that was driving that load,if no energy was lost in the process

I thought the answer to that was on that web site I posted.  You can change some concentrated pressure energy to a less concentrated pressure energy, and in the process use that transfer to do some work, to obtain some energy of a different form.  That "concentration" of pressure energy has some value and the fact that it has been reduced to a lower value is in itself an energy change.  I am not a thermodynamic engineer, I am a poor electronic engineer from the vacuum tube era (don't understand those pesky transistors) and a self taught physicist, so I have difficulty with words like enthalpy and entropy, but I am sure that one of those describes that "concentration".

The fallacy in your argument is "no energy being lost in the process".  According to your analysis you could follow up your two tank experiment with a four tank experiment where you again double the volume and extract energy.  You can go on doing this ad nauseam until you reach a miniscule pressure in an enormous volume and still carry on doing it.  But we know that pressure comes from molecular collisions and eventually you run out of molecules, so there is the flaw.  Your pressure energies, when transferred to say a piston movement that then creates a volume increase, yields infinite energy using our practical pressure-force v distance rules, so when you say the stored energy is the same you are saying infinite energy.  But it is not infinite, we run out of molecules.  So the simple rules concerning energy and pressure and fluid movement have limited application in the wider scheme of things, and we need some other measure of stored energy, not just pressure and volume, perhaps something like enthalpy or entropy.  In your simple experiment the change in apparent energy before and after may not even be measurable, but it is there.  And there is no going back while dealing with gas.  If there is a phase change (e.g. gas to liquid and vice versa) then the story is different.  Here endeth the first lesson.

Smudge
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee

I thought the answer to that was on that web site I posted.  You can change some concentrated pressure energy to a less concentrated pressure energy, and in the process use that transfer to do some work, to obtain some energy of a different form.  That "concentration" of pressure energy has some value and the fact that it has been reduced to a lower value is in itself an energy change.  I am not a thermodynamic engineer, I am a poor electronic engineer from the vacuum tube era (don't understand those pesky transistors) and a self taught physicist, so I have difficulty with words like enthalpy and entropy, but I am sure that one of those describes that "concentration".

The fallacy in your argument is "no energy being lost in the process".  According to your analysis you could follow up your two tank experiment with a four tank experiment where you again double the volume and extract energy.  You can go on doing this ad nauseam until you reach a miniscule pressure in an enormous volume and still carry on doing it.  But we know that pressure comes from molecular collisions and eventually you run out of molecules, so there is the flaw.  Your pressure energies, when transferred to say a piston movement that then creates a volume increase, yields infinite energy using our practical pressure-force v distance rules, so when you say the stored energy is the same you are saying infinite energy.  But it is not infinite, we run out of molecules.  So the simple rules concerning energy and pressure and fluid movement have limited application in the wider scheme of things, and we need some other measure of stored energy, not just pressure and volume, perhaps something like enthalpy or entropy.  In your simple experiment the change in apparent energy before and after may not even be measurable, but it is there.  And there is no going back while dealing with gas.  If there is a phase change (e.g. gas to liquid and vice versa) then the story is different.  Here endeth the first lesson.

Smudge

100 amps at 1 volt is as good as 100 volts at 1 amp.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Elite Experimentalist
Hero Member
*****

Posts: 1399
... .-.. .. -.. . .-.
It really is a fascinating subject.
Here's a real world use for such a thing as the water buckets.
Say, for example a farmer collects rain water to put in the troughs of cows in the milking shed. The water collection is in a large vessel and he never needs all of it. He also has a smaller tank from which to draw the water at a reasonable pressure with an attached hose....however, the troughs need to be filled first thing in the morning, in the dark. The system would spin a generator, make light and enable him to see what he's doing.
All for no electrical or water cost. 

The replacement part is the natural rain, that always used to fall into the larger vessel anyway.


---------------------------
ʎɐqǝ from pɹɐoqʎǝʞ a ʎnq ɹǝʌǝu
   

Group: Professor
Hero Member
*****

Posts: 1953
100 amps at 1 volt is as good as 100 volts at 1 amp.
And as good as 100 mega-volts at 1 micro-amp.  And as good as 1032 volts at 10-30 amps until you realize that 10-30 amps is a current of just one electron every 5078 years!  In other words there is a breakdown between our power=volts*amps laws because current comes in chunks called electrons and when the chunks get too far apart we can't talk of power as we know it.  Same with pressure that comes in molecular collision chunks.  Your 50L tank at 40 PSI has a certain energy, and although the product of pressure and volume is dimensionally energy that 50*40 (which would be a bastard unit) is not the energy available.  You can get far more than that.  With no losses your discharge of that 50L tank down to half pressure yields an energy of 50*40*loge(2) which is 0.69315 of that pressure*volume product.  Discharge that new 100L volume down to half pressure and you get another 0.69315 of that same pressure*volume product, and so on.  But you can't go on for ever doing that for reasons that I have outlined.  So there must be some other measure of the actual available energy in that original 50L at 40 PSI.  It is that peculiar huge value of energy remaining that changes when you do your initial discharge.  It is a reduction in available energy and IMO that explains where your gained energy came from.  This is not something taken from a text-book, this is just my reasoning.  It doesn't alter the fact that you got that 0.69315*P*V amount of energy and that seemingly you have ended up with the same P*V that appears to offer the same stored energy as you had originally.  I am saying that P*V is not the stored energy.  I can't tell you what the stored energy is but it is much greater than simply P*V.  50L at 40 PSI has greater stored energy than 100L at 20 PSI, and the reduction in stored energy accounts for the amount you gained.

Smudge
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 4727


Buy me some coffee
And as good as 100 mega-volts at 1 micro-amp.  And as good as 1032 volts at 10-30 amps until you realize that 10-30 amps is a current of just one electron every 5078 years!  In other words there is a breakdown between our power=volts*amps laws because current comes in chunks called electrons and when the chunks get too far apart we can't talk of power as we know it.  Same with pressure that comes in molecular collision chunks.  Your 50L tank at 40 PSI has a certain energy, and although the product of pressure and volume is dimensionally energy that 50*40 (which would be a bastard unit) is not the energy available.  You can get far more than that.  With no losses your discharge of that 50L tank down to half pressure yields an energy of 50*40*loge(2) which is 0.69315 of that pressure*volume product.  Discharge that new 100L volume down to half pressure and you get another 0.69315 of that same pressure*volume product, and so on.  But you can't go on for ever doing that for reasons that I have outlined.  So there must be some other measure of the actual available energy in that original 50L at 40 PSI.  It is that peculiar huge value of energy remaining that changes when you do your initial discharge.  It is a reduction in available energy and IMO that explains where your gained energy came from.  This is not something taken from a text-book, this is just my reasoning.  It doesn't alter the fact that you got that 0.69315*P*V amount of energy and that seemingly you have ended up with the same P*V that appears to offer the same stored energy as you had originally.  I am saying that P*V is not the stored energy.  I can't tell you what the stored energy is but it is much greater than simply P*V.  50L at 40 PSI has greater stored energy than 100L at 20 PSI, and the reduction in stored energy accounts for the amount you gained.

Smudge



The process i am showing here is an Isothermal process (as clearly stated in my first post),and in an isothermal process,the internal energy of the expanding system must remain a constant.
In other words,when the temperature is held at a constant,the internal energy remains the same.
As the pressure in tank A decreases,the temperature of the gas decreases,but as the pressure increases in tank B,the temperature of the gas increases.
What is lost in one,is gained in the other-->energy is conserved.

But regardless of that,you seem to be missing the point here.
What i am showing is,there is no change in outcome whether or not that gas dose work when flowing from one tank to the other.
It makes no difference whether or not that air motor/generator/load is there--the outcome is the same.

The other thing that you should be looking at,is the !now! 2 half full tanks of compressed gas can do the same amount of work as that of the one full tank of compressed gas--nothing is lost in the process,and so,work was done without loss of energy.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Pages: [1] 2
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2025-03-17, 05:40:13
Loading...