Open Systems

[tinman]

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.

N

ow, 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

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.

[tinman]

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.

[tinman]

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?

[profitis]

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.

[MarkE]

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 --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.