Open Systems

[tinman]

 

Where and when did I make such a claim?

Post 257-Quote: First, there is a reflection, but it is so small that for all practical purposes we neglect it.

So, the regulator does not guarantee pressure.  You need pressure gauges / sensors to know the pressure.

You must have a crappy regulator. Mine has a pressure gauge on both the supply and delivery side.Once the delivery side has reached the set value,gas discontinues to flow.The pressure holds stedy at the preset value.The pressure gauge on the delivery side of the regulator shows actual cylinder pressure. So the cylinder/ram dose have a pressure gauge,and the regulator will hold stedy at the set value of 50psi regardless of flow.

You need to know the pressure, temperature, volume and gas composition of both volumes before you open the valve.

This is just common sense,and it is why i never included this in the quick test sketchup.When calculating energy gains or losses,of course we measure what we have in the begining.

You need to know the pressure, temperature, volume and gas composition of both volumes immediately after you finish the stroke.

As i have stated before.__> this we know.

Regulators only regulate if there is a gas flow

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Incorrect. Regulators regulate maximum avaliable pressure.Once that set pressure has been reached within the ram,the gas will discontinue to flow.

A restriction only creates a pressure drop when there is flow.

We are not restricting flow,we are regualting pressure

That tells you the ending energy.  GIGO.  You didn't know the energy when you started

As stated above-this is common sense. Of course we calculate the avaliable energy at the start of the test. Spoon feeding is starting to come to mind here

You didn't determine the energy when you ended because you incorrectly assumed that the pressure was the regulator setting

You need to get your self a decent regulator that actually regulates set maximum pressure.

So go back to Square One, instrument so that you can determine the actual starting and ending energies and then see if you still think you observe work being done for free.

Already covered above. I really dont think spoon feeding is required ATM.

You are removing energy from the supply and putting it into the ram gas volume.

Indeed we are. And as the system is still closed,the energy loss is where? We have a 10 ltr bucket that has 5 ltr's of water in it.we tip 2 ltr's into another bucket-->how much water do we now have in the two buckets in total?.

Any value less than infinity removes energy from the source.

No,it dose not !remove! any energy,it transforms energy.

I am sorry, but you need to review your Kepler.

No thanks,i will stick with actual events and outcomes. Are you really saying that i will loose more gas energy in test 2,than i would in test 1 . Lol,think i'll leave the books with you on this one,and stick to what ihave observed over the last 13 years-->actual facts.

Setting up strawman arguments does not help your case.

And yet here you are,useing the !one fits all! approach. So i'll ask again-->please show the experimental data recorded around the system i present. 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.

[MarkE]

So now you don't understand the word "negligible"

Anyway so I take it we can look forward to your carefully obtained data that shows you can perform external work without losing that energy from the source gas volume.  Is that correct?

[tinman]

So now you don't understand the word "negligible"

I do. Do you accept the fact they you did indeed say there is a reflection.

Anyway so I take it we can look forward to your carefully obtained data that shows you can perform external work without losing that energy from the source gas volume.  Is that correct?

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.

[tinman]

The whole idea her is to show that secondary systems(open systems) can feed of the energy from the primary system with out there being any effect on the primary system-yes,no equal and opposite reaction.This is where those boundaries come into it,and they are set between the device that dose not reflect on the other,and the other device. One is coupled to the other,but the other is not coupled to the first-hence the solar panel and light experiment.

Have you ever put a light bulb inside a reflective ball(like a ball that has a mirror surface all around it's surface on the inside)and tried to cast a shadow inside that ball?.

[MarkE]

The whole idea her is to show that secondary systems(open systems) can feed of the energy from the primary system with out there being any effect on the primary system-yes,no equal and opposite reaction.

Do you understand that equal and opposite reaction:  Newton's Third Law is about force and not energy?  The equal and opposite reactions occur where material/energy leaves one thing, and then again where it impacts another.

This is where those boundaries come into it,and they are set between the device that dose not reflect on the other,and the other device. One is coupled to the other,but the other is not coupled to the first-hence the solar panel and light experiment.

Have you ever put a light bulb inside a reflective ball(like a ball that has a mirror surface all around it's surface on the inside)and tried to cast a shadow inside that ball?.

No, I can't say that I ever attempted to amuse myself by trying to do that.