Open Systems

[MarkE]

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.

So again just for rough checks:

You've got 120psig in a 20 liter pressure vessel, and you have a ram that you fill from empty with 750ml @ 50psig:

(120psig + 15psi ambient) * 20 liters => (X psig + 15psi ambient) * 20 liters + 750ml @ 65psi - 750ml @ 15psi

If no energy is added or removed, X <= (2700 - 37.5)/20 - 15 <= 118.13 psig.  If you are reading pressures higher than that, then something is very wrong.

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.

Why?  They are your tests.  I did not design them.

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?.

That would tell you that there is a gross misunderstanding in the test set-up, because at face value it would be claiming that performing work creates energy, which no one has ever previously observed.

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?.

If you perform work anywhere, that energy has to come from someplace.  So if you pump up capacitors with 1250J, then you need to take at least 1250J from someplace else.  If you are not measuring that loss, then barring a Nobel Prize headed your way, you have a measurement / accounting problem.

[tinman]

So again just for rough checks:

You've got 120psig in a 20 liter pressure vessel, and you have a ram that you fill from empty with 750ml @ 50psig:

(120psig + 15psi ambient) * 20 liters => (X psig + 15psi ambient) * 20 liters + 750ml @ 65psi - 750ml @ 15psi

If no energy is added or removed, X <= (2700 - 37.5)/20 - 15 <= 118.13 psig. 

So how much energy in joules did we loose from the 20 ltr supply vessel,and what do we now have in the 750ml cylider in joules of energy?-so as we can calculate the loss in total.
Cheers

[Pirate88179]

Tinman:

Congratulations on being selected the Experimenter Of The Month on Revolution Green.

Bill

[tinman]

Tinman:

Congratulations on being selected the Experimenter Of The Month on Revolution Green.

Bill

Ya what >?
Im not seeing it anywhere.

[Pirate88179]

Ya what >?

Sorry, I now see that this was in August of 2013.  Well, congratulations anyway...better late then never.

I stumbled across that story and assumed it was a current one.

Sorry,

Bill