Open Systems

[MarkE]

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:  U_{GAS_X1}
2) The ram moves from position X₁ to position X₂ against an opposing force:  F_X1 to F_X2.  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:  W_EXTERNAL. 
3) The energy in the gas after the stroke has been executed is:  U_{GAS_X2}.
4) You contend that:
a) U_{GAS_X2} > U_{GAS_X1}, AND
b) W_EXTERNAL > 0, AND therefore:
c) U_{GAS_X2} + W_EXTERNAL > U_{GAS_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.

[LibreEnergia]

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.

[tinman]

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?

[LibreEnergia]

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.

[MarkE]

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.