Open Systems

[tinman]

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.

[tinman]

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 n_{PV_START} molecules in the tank, then without any energy loss, the new pressure would be: 

P_{PV_END} = P_{PV_START}*n_{PV_START}/(n₁+n_{PV_START}), which is to say a difference of: 
P_{PV_START} - P_{PV_END} = P_{PV_START} * n₁/(n₁+n_{PV_START}).

Just for a rough estimate:  If P_{PV_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 = K_GAS*2.425 cu ft psi
n_{PV_START} = K_GAS*240 cu ft psi
P_{PV_END} = 120psi * 240/(242.425) = 118.8psi   ETA corrected from n_{PV_END}
P_{PV_START}  - P_{PV_END} ~= 1.2psi ETA corrected from n_{PV_START}  - n_{PV_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.

[MarkE]

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.

[MarkE]

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.

[tinman]

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?.