Open Systems

[tinman]

Gamma for the nitrogen and oxygen which makes up most of the air is 1.4.  So:

P*V^1.4 = K.

Whereas the energy is P*V.

Consequently, for a change in volume from V_START to V_END the remaining energy as a ratio to the starting energy is:

E_END/E_START = P_START*(V_START^1.4/V_END^0.4)/(P_START*V_START)

P_START appears in the numerator and the denominator and cancels, and V_START is raised to the 1.4th power in the numerator and the 1.0th power in the denominator, so simplifying we get the remaining energy percentage as:

E_END/E_START = (V_START/V_END)^0.4 and the percentage energy lost is:

E_{INTERNAL_LOSS}/E_START = (V_END^0.4 - V_START^0.4)/V_END^0.4

I'll have to go back and look for the dimensions to plug in values for the starting and ending volumes.  For the first approximation we would just use the pressure vessel volume for V_START and the sum of the cylinder extension volume plus the pressure vessel volume for V_END.

For 20 liter pressure vessel, and a 750ml ram, the approximate percentage loss to heat on each stroke is:

E_LOSS% = (20.75^0.4 - 20^0.4 )/20.75^0.4 = 1.46%

The initial energy is 120psi * 20 liters =  6895 P/psi * 120 psi * 0.02m³ ~= 16.5kJ.  1.46% * 1.65kJ ~= 242J internal energy loss per stroke.  This is approximate because we are treating the ram cylinder volume as completely void, when it at 1 ATM.  This rough approximation s/b within 5-10% of the actual loss.  If you vent the ram to reset it, then after about 50 operations the energy in the pressure vessel and the pressure should both be about 50% of the starting value.

So to get this straight,each stroke we loose heat?,and to where-the enviroment?
And next-->this heat loss results in a drop of preasure.?

So you are sure there is no way we could gain energy from the enviroment through the open part of the system(other than heat)/.

I have all the equipment now,and we are going to change things around a little to make the outcome much easyer to measure. We will be adding a second tank to capture this !extra! energy.

I will draw up a schematic when i get back from my next freight trip-should have it here by saterday arvo.

[MarkE]

Each stroke you lose internal energy in the gas to heat that then eventually finds its way out of the system.  You have practical experience with rams and know they heat up as they operate.  You have practical experience with aerosol cans and know that they cool down.

[tinman]

Each stroke you lose internal energy in the gas to heat that then eventually finds its way out of the system.  You have practical experience with rams and know they heat up as they operate.  You have practical experience with aerosol cans and know that they cool down.

The thing with the aerosol can is you have a liquid turning to gas when the gas is released from the can-same as the tap freezing on a LPG bottle when you open the valve. This rapid expansion is what creates the cooling effect. But what if the expansion is gas expanding slowly,and in the other vessel the gas is being compressed at a faster rate?.

Tomorrow i will liquid measure the two tanks,and post a diagram of how the first test setup will be.
What i will need to know after i have the tank capacities,is what pressures are needed in each tank to be a unity measurement.

So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.

The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.

More info tomorrow.

[MarkE]

The thing with the aerosol can is you have a liquid turning to gas when the gas is released from the can-same as the tap freezing on a LPG bottle when you open the valve. This rapid expansion is what creates the cooling effect. But what if the expansion is gas expanding slowly,and in the other vessel the gas is being compressed at a faster rate?.

The change in internal energy depends on the degrees of freedom of the molecules and the change in volume.  How fast you change the volume determines the power.

Tomorrow i will liquid measure the two tanks,and post a diagram of how the first test setup will be.
What i will need to know after i have the tank capacities,is what pressures are needed in each tank to be a unity measurement.

So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.

The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.

More info tomorrow.

If you use a small tank as your reservoir raising the pressure in a larger vessel then you will lose a greater percentage of energy with each operation than if you use a large tank as the high pressure reservoir.  Remember that when you calculate energy you need to use absolute temperature and pressure values.

[telecom]

So to understand the basics of the test,the two tanks will be joined via a ball valve. We will fill the small tank to a pressure of say 40psi gauge pressure(all measurements will be psi and gauge pressure). We then will open the ball valve so as the gas from the small tank will flow into the large tank until equilibrium has been reached. Once this test has been done 3 or 4 times,and a average measurement has been reached,then i will show how that average value can be lifted without any work from the outside being needed to do so. I will show how the energy within the two vessels can be raised simply by using the energy already stored within the vessel that is filled with our gas at the start of each test. The two vessels are our closed system-in that no pressurised gas can escape the system. Our open system is the atmosphere around the two vessel's. This extra energy that will be entering the system is not heat energy-and that is all we will say about that ATM. This system has no ram-that system will come later. The first system is very simple,and is just a proof of concept.

The aim ATM is to do nothing more than to raise the pressure and the (should be) temperature over that of what would be a normal outcome-the values reached in the first 3 or 4 test runs without anything attached to the system.

More info tomorrow.

This sounds very interesting!
Looking forward for the results.
Regards