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Author Topic: Open Systems  (Read 107753 times)

Offline MarkE

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Re: Open Systems
« Reply #75 on: February 01, 2015, 11:15:10 PM »
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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Re: Open Systems
« Reply #75 on: February 01, 2015, 11:15:10 PM »

Offline pomodoro

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Re: Open Systems
« Reply #76 on: February 02, 2015, 01:33:57 AM »
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.

Offline tinman

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Re: Open Systems
« Reply #77 on: February 02, 2015, 06:14:19 AM »
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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Re: Open Systems
« Reply #77 on: February 02, 2015, 06:14:19 AM »
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Offline profitis

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Re: Open Systems
« Reply #78 on: February 02, 2015, 06:51:01 AM »
Pomodoro.who was taxed in the compression cycle: tinman exclusivey or tinman plus environment.

Offline tinman

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Re: Open Systems
« Reply #79 on: February 02, 2015, 07:58:45 AM »
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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Re: Open Systems
« Reply #79 on: February 02, 2015, 07:58:45 AM »
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Offline minnie

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Re: Open Systems
« Reply #80 on: February 02, 2015, 09:00:00 AM »



       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.

Offline MarkE

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Re: Open Systems
« Reply #81 on: February 02, 2015, 09:01:08 AM »
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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Re: Open Systems
« Reply #81 on: February 02, 2015, 09:01:08 AM »
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Offline MileHigh

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Re: Open Systems
« Reply #82 on: February 02, 2015, 10:29:14 AM »
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

Offline MileHigh

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Re: Open Systems
« Reply #83 on: February 02, 2015, 10:52:17 AM »
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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Re: Open Systems
« Reply #83 on: February 02, 2015, 10:52:17 AM »
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Offline pomodoro

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Re: Open Systems
« Reply #84 on: February 02, 2015, 11:20:27 AM »
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?

Offline LibreEnergia

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Re: Open Systems
« Reply #85 on: February 02, 2015, 11:43:21 AM »
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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Re: Open Systems
« Reply #85 on: February 02, 2015, 11:43:21 AM »
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Offline tinman

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Re: Open Systems
« Reply #86 on: February 02, 2015, 01:33:48 PM »
  I might be blind but how does the environment give energy to this system?
Quote
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.

Offline tinman

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Re: Open Systems
« Reply #87 on: February 02, 2015, 01:40:47 PM »
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.

Offline pomodoro

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Re: Open Systems
« Reply #88 on: February 02, 2015, 02:19:28 PM »
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?

Offline MarkE

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Re: Open Systems
« Reply #89 on: February 02, 2015, 02:52:45 PM »
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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