<<< And Mag's was correct. You cannot !loose! energy--it just dose not disappear--you cannot destroy energy--->basic 001 >>>

You are just one never-ending shameless barrel of monkeys Brad.  Did you ever hear the expression, "Energy lost to heat?"  Keep on being that creepy horrible nightmare from hell.  What ugliness there is under your skin.  Dunning-Kruger effect, OCD for being wrong or perceived as being wrong, and a creepy ugly bogan to boot.

Nice read...and yes it is odd how fast it can be read .

perhaps I might interject a thought regarding the Airtank and MH's analogy .

all venturi effects or Temp/Phase /gravitational/ Buoyancy  effects out of the equation .


if you place two identical tanks next to each other ,

tank A has 100 lbs of air

Tank B Zero

you will never get over 50 lbs into tank B from Tank A

So a total transfer is impossible , but all the energy is still in the system.

You can if you use the air from tank A to drive a turbine which runs a generator which runs a compressor...you still have the air, and it is compressed again...well...not all of it of course...but more than 50%.

Bill

Webby, pull yourself together and show some self-respect.

Brad, no, if you are going to draw an analogy between capacitors and air tanks then you wait for the air tanks to reach thermal equilibrium and you have lost half of your energy.

Both of you are being ridiculous and making fools of yourself.  That's no surprise.

Nice read...and yes it is odd how fast it can be read .

perhaps I might interject a thought regarding the Airtank and MH's analogy .

all venturi effects or Temp/Phase /gravitational/ Buoyancy  effects out of the equation .


if you place two identical tanks next to each other ,

tank A has 100 lbs of air

Tank B Zero

you will never get over 50 lbs into tank B from Tank A

So a total transfer is impossible , but all the energy is still in the system.

Absolutely correct.
Now,what if that transfer line between tank A and tank B had a venturi entry into it.
Tank A starts with 100 psi,tank B starts at atmospheric pressure-0 gauge pressure.
We open the valve on tank A to allow the compressed air to start to flow into tank B,and at the same time it is drawing in ambient air through the venturi. At the point where the pressure in tank B stop's the venturi effect working,the venturi is shut off,so as air cannot escape via the venturi valve. The compressed air is allowed to continue to flow into tank B,until pressure equilibrium is reached between the two tanks.
The result is
1-you finish with a higher equilibrium pressure in the two tanks,as you added more volume of air into the system via the venturi.
2-As the pressure is now higher,so will be the two end temperatures in each tank.
3-Because you now have a higher pressure equilibrium in the two tanks,the volume of compressed air has also increased.

This leaves us with more stored potential energy that we started with-and that's a fact.


Brad

Webby, pull yourself together and show some self-respect.

Brad, no, if you are going to draw an analogy between capacitors and air tanks then you wait for the air tanks to reach thermal equilibrium and you have lost half of your energy.

.

You are making a fool of your self MH.

When i did the test with MarkE,one tank was pressurized to a certain pressure,and then left to sit to reach room temperature. This meant a drop in pressure and inside gas temperature.
The tank was then topped up to our regulated pressure (set pressure throughout the tests).
Once again,it was left to drop to room temperature. The compressed gas in tank A was then let flow into tank B,and then once pressure equilibrium was reached,the gas inside the tanks was once again left to sit until room temperature was reached.

With the venturi switched off,we had an exact potential energy transfer from one tank to both tanks--nothing was lost. This set the bench mark for the venturi tests,and also made it clear that the tests were being carried out in a controlled manner.

When the same tests were carried out with the venturi in play,the results showed an increase of 16% of stored potential energy in the two tanks.

The very same tests were carried out with all of the DUT now heavily insulated in thermal wool.
This allowed us to skip the waiting for the gas inside the tanks to drop to room temperature.
The results were exactly that of the first set of tests.

So do not tell me that energy is lost through the air tank transfer situation,as you are totally wrong.

Both of you are being ridiculous and making fools of yourself.  That's no surprise

What is no surprise MH,is you got it wrong again.
This is becoming a habit with you.
Time and time again,you have made claims that i am wrong,and time and time again,i have proven you to be wrong-->and it just happened again.


Brad

Brad:

It's pure sleaze for you to pose a question and then pull a bait and switch where you draw extra air into the destination tank with a venturi without stating that in the first place.  You have no shame and you were too lazy or sleazy to even define the question properly.

And I will concede that you are most likely correct about the standard air tank test and I am wrong.  This is a setup where the environment itself exchanges energy both ways with the two air tanks whereas for two capacitors there is a one-way exchange of energy with the environment, the heat is considered lost.  When the pressurized tank is discharged into the empty air tank, energy is lost to heat in the valve, the pressurized tank does work on the unpressurized tank, the pressurized tank draws heat from the environment, and the unpressurized tank starts to put heat into the environment.  So this is a thermodynamic problem and accounting for everything is pretty tricky when this is not your forte (like me).  I was silly by forgetting about the thermodynamic angle and the energy exchange with the environment.

<<< When the same tests were carried out with the venturi in play,the results showed an increase of 16% of stored potential energy in the two tanks. >>>

Perhaps, but now you are playing in the big leagues and you can't cherry pick if you are going to be thorough.  Energy had to be taken out of the external environment to do that, so there is no real energy gain.

I am not even going to try to work out the specifics for the two tanks and will take your word for it.  I would not even consider the venturi example, it's just a stupid bait and switch on your part.

<<< Time and time again,you have made claims that i am wrong,and time and time again,i have proven you to be wrong-->and it just happened again. >>>

It just happened again, yes.  But you seriously would not want to have an independent audit of errors in this thread made by both of us.  Because of your OCD, your head would explode if all of your errors were pointed out to you.  For Christ's sake, you read your own quote with a jarring spelling mistake and five minutes later you forgot that fact or it didn't even register in your brain that you had read your own quote.  So you end up accusing me of your own spelling mistake - five minutes after you read your own mistake!

<<< No MH--i defended myself against your lies--and you know it. >>>

I did not lie and it's not my fault that your brain cannot process information properly and understand what I stated in my posting.

MileHigh

   Have I got it? You transfer from one tank to another and end up with say
   66% of your original pressure so in fact you're only losing 33% by doing
   so. No overunity power supply here!!
             John.