Open Systems

[MarkE]

Ok,i have just messured the two tank's,and i was out by double lol.
The small tank is exactly 10 ltr's,and the big tank is exactly 20 ltr's.

There is a reason i will be using the small tank for the pressurized tank,and that will become apparent later on in the testing. The two tanks will be joined via a ball valve that will be opened to let the pressurized gas drain into the large tank. Each tank will have it's own temperature gauge,but only the small tank will need a pressure gauge-as the pressure at the end of each test will be the same in both tanks.

So a quick bit of calculating for you Mark,if you could please.
Our small tank(10LTR) will have a pressure of 40psi gauge pressure at the start of the test,and the large tank will have 1 ATM. The temperature will be close to 25*C-avaerage ambiant temperature here at the moment-->will give exact temperature on day of test. But going on these measurement's,can you calculate what we should have in the way of temperature in each tank,and the pressure in both tanks?.

Roughly:
55 psi * 10^1.4 liters +
15 psi * 20^1.4 liters
=  P_EQUALIZED * 30^1.4 liters
P_EQUALIZED = ((55*25.1) + (15*66.3))/116.9 = 20.3 psi = 5.3psi gauge

Applying the ideal gas law:  PV = nRT, T = PV/(nR).  T_END/T_START = (PV)_END/(PV)_START * n_START/n_END

The small tank PV product at the end K*20.3psi*10l instead of K*55psi*10l.
So (PV)_END/(PV)_START = 20.3/55 = 0.369.

The proportion of molecules in the small tank changes from 55*10/(55*10 + 15*20) to 10/30. 
So, n_START/n_END = 550/850 * 3/1 = 1.941.

The absolute temperature then is:  T_END = 0.369 * 1.941 * T_START = 0.716 * T_START. 
If T_START = 25C = 298.2K then T_END = 214K = -59C.

The large tank PV product at the end K*20.3psi*20l instead of K*15psi*20l.
So (PV)_END/(PV)_START = 406/300 = 1.353.

The proportion of molecules in the large tank changes from 15*20/(55*10 + 15*20) to 20/30. 
So, n_START/n_END = 300/850 * 3/2 = 0.529.

The absolute temperature then is:  T_END = 1.353 * 0.529 * T_START = 0.716 * T_START. 
If T_START = 25C = 298.2K then T_END = 214K = -59C.

I am a bit skeptical that the predicted equilibrium temperature is so low, but the work is there to see. 

Anyway, this gives us a large separation from what constant energy predicts, which is that you would get ((10*(15+40)) + (20*15))/30 - 15 = 10*40/30 = 13.3psi gauge.  And since under the constant energy premise: the PV product is constant as is the number of molecules there would be no temperature change.

So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.

[LibreEnergia]

So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.

I'd predict you will see the gauge pressure of 13.3 psi  fairly quickly. What will happen is heat will flow from ambient surroundings into the combined cylinders as the gas expansion causes them to become cooler. The rate of change of temperature can't be calculated without knowing how well insulated they are and would be described by differential equations.

You would only see the lower pressure value if the cylinders are perfectly insulated.

[MarkE]

I'd predict you will see the gauge pressure of 13.3 psi  fairly quickly. What will happen is heat will flow from ambient surroundings into the combined cylinders as the gas expansion causes them to become cooler. The rate of change of temperature can't be calculated without knowing how well insulated they are and would be described by differential equations.

You would only see the lower pressure value if the cylinders are perfectly insulated.

Because the tanks are pretty reflective most of the transfer will be conductive. The thermal resistance of the gas is pretty high.  I expect that the time constant is at least a couple minutes.   That's plenty of time for tinman to take his pressure and temperature readings.  Even as the system moves on him, he can get a decent qualitative result.  Any pressure well below 13.3psi and any temperature well below ambient should be adequate proof to him that the PV product is not constant.  If he wants he can later pack the whole thing in some decent insulation to show himself how that slows down the return to room temperature.

[tinman]

Roughly:


I am a bit skeptical that the predicted equilibrium temperature is so low, but the work is there to see. 

Anyway, this gives us a large separation from what constant energy predicts, which is that you would get ((10*(15+40)) + (20*15))/30 - 15 = 10*40/30 = 13.3psi gauge.  And since under the constant energy premise: the PV product is constant as is the number of molecules there would be no temperature change.

So, if you see an equalized net gauge pressure of ~13.3psi instead of a much lower pressure of ~5.3psi then things will be interesting.

So the first thing i need to do is run say 5(what we will call) open test's. This will give us a figure to work with,and one to beat. These test will be just opening the ball valve between the two tank's,and let the gas flow from one to another until an equilibrium is reached between the two tank's. If i record the start pressure in the small tank(the other will be 1 ATM of corse),and the temperature in both(as well as ambiant temperature),then open the valve to allow gas flow,then record the end pressure in both tanks-along with end temperature in both tank's as soon as gas flow has stopped,then this will give us close to an instantaneous measurement. I will then leave the system rest for a period of 2 minutes,and once again record the values. This will give us an indication as to how much the enviroment is contributing in the way of heat energy within that 2 minutes.

When i apply my first system,we will carry out the tests again as we did above. With the first system in opperation,it will take about 2 minutes to complete one cycle-->pressure in both tanks reach equilibrium.

Regardless of what LE says about drawing heat in from the enviroment,the heat of the gas can never exceed that of the outside enviromental temperature-->if it dose,we then have free energy.
The goal is to raise the pressure higher than the 13.3psi Mark has stated. But first we must see if that is indeed what we will have when the primary test are carried out on the system.

[MarkE]

So the first thing i need to do is run say 5(what we will call) open test's. This will give us a figure to work with,and one to beat. These test will be just opening the ball valve between the two tank's,and let the gas flow from one to another until an equilibrium is reached between the two tank's. If i record the start pressure in the small tank(the other will be 1 ATM of corse),and the temperature in both(as well as ambiant temperature),then open the valve to allow gas flow,then record the end pressure in both tanks-along with end temperature in both tank's as soon as gas flow has stopped,then this will give us close to an instantaneous measurement. I will then leave the system rest for a period of 2 minutes,and once again record the values. This will give us an indication as to how much the enviroment is contributing in the way of heat energy within that 2 minutes.

When i apply my first system,we will carry out the tests again as we did above. With the first system in opperation,it will take about 2 minutes to complete one cycle-->pressure in both tanks reach equilibrium.

Regardless of what LE says about drawing heat in from the enviroment,the heat of the gas can never exceed that of the outside enviromental temperature-->if it dose,we then have free energy.
The goal is to raise the pressure higher than the 13.3psi Mark has stated. But first we must see if that is indeed what we will have when the primary test are carried out on the system.

LibreEnergie is not suggesting that the environment is adding energy to the tanks over and above what was originally there.  When the valve opens and the gasses are moving the small more pressurized tank temperature drops, while the bigger tank at 1 ATM heats rejecting heat to the environment.  If you put the whole thing in one really well insulated container then as the tanks equalize, their temperatures will also equalize getting much colder in the process.  It takes heat from outside to restore the total energy seen as the sum total PV of the combined tanks.

The longer it takes the tanks to equalize the less sever the temperature and pressure drop will be.  By the seat of my pants a couple of minutes is probably short enough that you will see a significant temperature drop and pressure shortfall versus the constant energy / constant PV produce hypothesis that you are testing.