Open Systems

[MarkE]

Thanks.  I want to know the weight of the tanks themselves empty.  I assume they are steel.  This would give me a good indication of their thermal capacity.  The two things that I have not taken into account yet is the amount of heat that the tanks can store in their walls and Van der Waals forces.

If for even a rapid:  say 5 second or 10 second equalization you are getting back to a pressure indicative of constant energy then the tanks are recuperating for you as big heat sinks / sources.  The good, bad, and indifferent about that is that if your target machine has similar geometries and you operate at a low rate then you won't end up losing much energy just to moving gas from your main vessel.  The primary energy that you lose from within the combined pressure vessel and ram will be the work the ram performs on the outside world.

[tinman]

Thanks.  I want to know the weight of the tanks themselves empty.  I assume they are steel.  This would give me a good indication of their thermal capacity.  The two things that I have not taken into account yet is the amount of heat that the tanks can store in their walls and Van der Waals forces.

If for even a rapid:  say 5 second or 10 second equalization you are getting back to a pressure indicative of constant energy then the tanks are recuperating for you as big heat sinks / sources.  The good, bad, and indifferent about that is that if your target machine has similar geometries and you operate at a low rate then you won't end up losing much energy just to moving gas from your main vessel.  The primary energy that you lose from within the combined pressure vessel and ram will be the work the ram performs on the outside world.

I will have to disconect the plumbing to weigh the tank's. I will do this tomorrow when i make the carry frame for the tanks. The mounting brackets have a rubber band that go's between them and the tanks,so that should provide some insulating properties between tank and mounting brackets. The brackets are also only an inch wide,so little surface contact anyway.

OK,the results of the tests you asked for Mark.
I filled tank A with 45psi in the hope that when it dropped down to ambiant temperature,it would be close to 40psi. Ambiant temperature at the time of test was 20*C,and it took about 5 minutes for the temperature of the gas in tank A to drop to that temperature. The pressure ended up at 40.4psi-so was preaty close.

Starting pressure tank A 40.4psi gauge pressure.
Temperature of both tabks was 20*C
Tank B=1 ATM.
End result was
Tanks A&B end pressure= 13.1psi
Tank A instant end temperature was 17*C
Tank B instant end temperature was 23*C

After 5 minutes sitting.
Tanks A&B pressure still 13.1psi
End temperature for both tanks was of course 20*C

I would also like you to crunch these numbers for me if you could Mark.
These are the average results over 4 runs with my first system enabled

Start of test was as above.
Tank A start pressure = 40.1psi
Tank A&B start temperature was 20*C.

End results.
Tank A&B end pressure = 16.2 psi
Tank A&B end temperature is of course 20*C.<--- This is NOT the ram test results. This is a much simpler design.

[MarkE]

I will have to disconect the plumbing to weigh the tank's. I will do this tomorrow when i make the carry frame for the tanks. The mounting brackets have a rubber band that go's between them and the tanks,so that should provide some insulating properties between tank and mounting brackets. The brackets are also only an inch wide,so little surface contact anyway.

OK,the results of the tests you asked for Mark.
I filled tank A with 45psi in the hope that when it dropped down to ambiant temperature,it would be close to 40psi. Ambiant temperature at the time of test was 20*C,and it took about 5 minutes for the temperature of the gas in tank A to drop to that temperature. The pressure ended up at 40.4psi-so was preaty close.

Starting pressure tank A 40.4psi gauge pressure.
Temperature of both tabks was 20*C
Tank B=1 ATM.
End result was
Tanks A&B end pressure= 13.1psi
Tank A instant end temperature was 17*C
Tank B instant end temperature was 23*C

After 5 minutes sitting.
Tanks A&B pressure still 13.1psi
End temperature for both tanks was of course 20*C

I would also like you to crunch these numbers for me if you could Mark.
These are the average results over 4 runs with my first system enabled

Start of test was as above.
Tank A start pressure = 40.1psi
Tank A&B start temperature was 20*C.

End results.
Tank A&B end pressure = 16.2 psi
Tank A&B end temperature is of course 20*C.<--- This is NOT the ram test results. This is a much simpler design.

ATM is assumed 101.3E3P
Start
Tank A .01m³ 278.6E3P gauge 379.9E3P abs 293.2K
Tank B .02m³ 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.9

End
Tank A .01m³ 90.3E3P gauge 191.6E3P abs 290.2K 
Tank B .02m³ 90.3E3P gauge 191.6E3P abs 296.2K
sum(PV/T) = 19.5

Energy instant end/Energy instant start = 19.5/19.9 = 98%

The second test data does not make sense.   It suggests accuracy issues in the measurements.

Start
Tank A .01m³ 276.5E3P gauge 377.8E3P abs 293.2K
Tank B .02m³ 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.8

End
Tank A .01m³ 111.7E3P gauge 213.0E3P abs 293.2K 
Tank B .02m³ 111.7E3P gauge 213.0E3P abs 293.2K
sum(PV/T) = 21.8

Energy instant end/Energy instant start = 21.8/19.8 = 110%

[tinman]

Energy instant end/Energy instant start = 19.5/19.9 = 98%

The second test data does not make sense.   It suggests accuracy issues in the measurements.
Start
Tank A .01m³ 276.5E3P gauge 377.8E3P abs 293.2K
Tank B .02m³ 0 gauge 101.3E3P abs 293.2K
sum(PV/T) = 19.8

End
Tank A .01m³ 111.7E3P gauge 213.0E3P abs 293.2K 
Tank B .02m³ 111.7E3P gauge 213.0E3P abs 293.2K
sum(PV/T) = 21.8

Energy instant end/Energy instant start = 21.8/19.8 = 110%

Do you find it odd that it is assumed that the tests were carried out correctly when an under 100% figure was achieved,and yet when an over 100% figure was achieved,then there must have been measurement error-->even though the tests were carried out in the very same way,with the very same equipment,in identical conditions.

All one has to do Mark to achieve this,is to introduce more gas(air) from the enviroment-->the open part of the system. You simply use the energy avaliable in tank A to pull in more gas from the surroundings of the device. If you think hard enough,you will be able to work out how you can draw more gas into the system without letting any of the gas contained within the system escape. Think along the lines of an gas diode,and you are on the right track. You will also then be able to work out why i chose the small tank as the energy storage tank,and the large tank as the collector. By using a very small jet,we can achieve a high speed air flow at low flow rates. It is this that is needed to achieve the effect you have seen in the results i posted. You must agree that even the smallest amount of gas introduced into the system from the enviroment,will only add to the end resulting energy amount.

Im not sure if you are able to do this calculation,but can you derive as to how much extra gas would be needed to be introduced to the system to achieve a result of 110%-eg-1 ltr at ATM ?.

[MarkE]

Do you find it odd that it is assumed that the tests were carried out correctly when an under 100% figure was achieved,and yet when an over 100% figure was achieved,then there must have been measurement error-->even though the tests were carried out in the very same way,with the very same equipment,in identical conditions.

Test results that at least superficially agree with 200+ years of thermodynamics pass sanity checks.  Tests that don't have that 200+ year hill of experience to overcome.

All one has to do Mark to achieve this,is to introduce more gas(air) from the enviroment-->the open part of the system. You simply use the energy avaliable in tank A to pull in more gas from the surroundings of the device. If you think hard enough,you will be able to work out how you can draw more gas into the system without letting any of the gas contained within the system escape. Think along the lines of an gas diode,and you are on the right track. You will also then be able to work out why i chose the small tank as the energy storage tank,and the large tank as the collector. By using a very small jet,we can achieve a high speed air flow at low flow rates. It is this that is needed to achieve the effect you have seen in the results i posted. You must agree that even the smallest amount of gas introduced into the system from the enviroment,will only add to the end resulting energy amount.

If you add energy by doing work then what is special?  At this point the most basic part of the rig is still being shaken out.  If the tests are as represented then the second set data is not reasonable.  If there is something you have not disclosed about the second set of tests, then having me run calculations on faulty presumptions is bound to yield GIGO results.

Im not sure if you are able to do this calculation,but can you derive as to how much extra gas would be needed to be introduced to the system to achieve a result of 110%-eg-1 ltr at ATM ?.

The question does not make any sense to me in the context of your two fixed volume closed container set-up.