Open Systems

[MarkE]

Absolute rubbish.
The work done to raise the pressure in tank A IS the stored energy in tank A. The efficiency of the compressor to perform this work has nothing at all to do with the efficiency of the DUT.

Wrong.
The numbers Mark crunched on the DUT fitted with the venturi show's a 10% increase of energy above that of the starting value. This stored energy can now perform more usful work than that of the energy amount we started with.

Well the big problem here is entropy.  You have managed to get the external system to add energy to your apparatus.  The problem is how much useful work you can do, and how much work you will have to do in order to get back to your starting condition.

Are you kiding me -->What next
Here you say it is possable to draw in extra energy from the enviroment,and then you ask !!Then what!!
Then we have more energy than what we started with. With this extra energy now avaliable,do you not think it possable that tank A can now be restored to it's original starting pressure,and we still have some stored energy left over in tank B?-->or dose this extra energy just up and vanish?

Well since you think you can, then all you need to do now is to try and figure a way to do that.

See above statement.

It's not meaningless at all. There is no point in going further until we have an increase in energy. This is like asking how fast the car go's before you know if the engine run's.

Absolute energy is not the only issue at hand.  Entropy is a big deal.

[tinman]

Well the big problem here is entropy.  . .Well since you think you can, then all you need to do now is to try and figure a way to do that.Absolute energy is not the only issue at hand.  Entropy is a big deal.

Well the big problem here is entropy.

And why is entropy such a big issue? ,when the  heat is both transferd into(entropy increases) and then disipated out of(entropy decreases) the system. The end temperature will always be the ambiant temperature in which the unit opperates.

The problem is how much useful work you can do, and how much work you will have to do in order to get back to your starting condition

If you would crunch those last numbers for me,i will have a starting point to work with. Im not asking you to verify the workings of some wonder machine(that bit will be up to me to do),i am asking for nothing more than the outcome or product of some results i have given.

.Well since you think you can, then all you need to do now is to try and figure a way to do that.

The way is figured out,but no more money will be spent until i know what i have in way of added energy. Im happy to spend the time and money,but i ask for help with the energy calculation's.

You have managed to get the external system to add energy to your apparatus

Yes,i have. But i need to know how much energy was added to the system before i can go any further. If you do not wish to do these calculations for me,then im sure there are others that will that reside in a different place. I have managed to amplify a given amount of energy at no cost to the original energy that created it. We now have a battery that self charges,a system that take a certain amount of energy,and increases it without loss. How can this not do more work than the original amount of energy we started with?. I can !without a doubt! show you the venturi effect increase a systems energy output while continually running,without any additional energy input-->would this be of any use?.The calculated input may be say 10 joules per second,while the output is say 15 joules per second. I will NOT however be held accountable for other devices inefficiencies-like a compressor.A compressor is not the DUT here,the same as a battery is not a battery charger.

[MarkE]

And why is entropy such a big issue? ,when the  heat is both transferd into(entropy increases) and then disipated out of(entropy decreases) the system. The end temperature will always be the ambiant temperature in which the unit opperates.

You will find out soon enough when you try to put all those molecules back in Pandora's 10 liter cylinder.

If you would crunch those last numbers for me,i will have a starting point to work with. Im not asking you to verify the workings of some wonder machine(that bit will be up to me to do),i am asking for nothing more than the outcome or product of some results i have given.

Evaluating numbers is senseless if the apparatus involved is not specified.  We've already been through that once.  Describe the machine accurately and I will run numbers.

The way is figured out,but no more money will be spent until i know what i have in way of added energy. Im happy to spend the time and money,but i ask for help with the energy calculation's.

Yes,i have. But i need to know how much energy was added to the system before i can go any further. If you do not wish to do these calculations for me,then im sure there are others that will that reside in a different place. I have managed to amplify a given amount of energy at no cost to the original energy that created it. We now have a battery that self charges,a system that take a certain amount of energy,and increases it without loss. How can this not do more work than the original amount of energy we started with?. I can !without a doubt! show you the venturi effect increase a systems energy output while continually running,without any additional energy input-->would this be of any use?.The calculated input may be say 10 joules per second,while the output is say 15 joules per second. I will NOT however be held accountable for other devices inefficiencies-like a compressor.A compressor is not the DUT here,the same as a battery is not a battery charger.

Well what has happened is energy has been moved (as LE pointed out that's something that a heat pump does), and the entropy has been increased.  There have been many people who thought that they could manipulate a heat pump:  Device that expends useful work moving a quantity of heat energy between two reservoirs into a free energy machine.  They run up against Sadi's Curse every time. 

The percentage heat that can be obtained from a reservoir is the ratio of the temperature change incurred reducing the temperature by conveying work to the hot temperature.  Because of that little formula PV = nRT, we know:

R is a constant, and if n doesn't change, then
PV = K₁T.  Consequently, as you increase the volume occupied by the gasses even with the same total energy, the usable energy that you can extract as work keeps shrinking.  So somewhat analagous to a water ram where we can pump a small fraction of water above the starting level by dumping a whole bunch of water to a lower level, you have with the venturi and check valve pumped some of the surrounding atmospheric air up to a higher pressure at the cost of dumping a bunch of air from the 10 liter vessel to a much lower pressure. 

[tinman]

 


R is a constant, and if n doesn't change, then
PV = K₁T. 

You will find out soon enough when you try to put all those molecules back in Pandora's 10 liter cylinder.

There comes a time when the energy drawn in from the enviroment surpasses that lost to entropy.

There have been many people who thought that they could manipulate a heat pump:

Yes,i know. I have been involved in many different variations of heat pump devices,and my system is represents none of them.

Consequently, as you increase the volume occupied by the gasses even with the same total energy, the usable energy that you can extract as work keeps shrinking.

Yes,this we know. But as i stated above,there is a point where the energy gathered by the systems opperation surpasses the losses.

So somewhat analagous to a water ram where we can pump a small fraction of water above the starting level by dumping a whole bunch of water to a lower level, you have with the venturi and check valve pumped some of the surrounding atmospheric air up to a higher pressure at the cost of dumping a bunch of air from the 10 liter vessel to a much lower pressure.

Yes,the pressure is much lower,but the volume has increased by 3X. If we had some calculations done,we would then know if the now lower pressure but increased volume can do more work than that of the higher pressure but lower volume can do. We now have a 30 ltr tank with a pressure of 18psi+,when at the start we had a 10ltr tank with a pressure of 40.2psi. Which configuration could supply a device that needs a constant pressure of say 2psi at a fixed flow rate of say 1ltr a minute to opperate(do work) for a longer period of time?.Is it 10ltr's at 40psi,or 30ltr's at 18psi?.

Evaluating numbers is senseless if the apparatus involved is not specified.  We've already been through that once.  Describe the machine accurately and I will run numbers.

well it then seems we are done here,as like you said,we have been through this before,and i thought i made it very clear that the device and it's workings would not be disclosed until such time as i am ready to do so. I ask nothing more that for you to calculate start and end energies in both tank's,and for some reason you need to know exactly how the device is set up.
I will try once again-->we have our two tank's,and gas is transfered from one tank to another via way of the ram that is doing work as this transfer takes place. The ram starts and finishes each cycle empty,and as in the venturi setup,the ram is also drawing gas in from the enviroment,and pumping it into tank B. The venturi is only capable of raising the pressure in tank B so much,and then it can no longer contribute to the system. The ram however has the ability to lift the pressure to a much higher level in tank B before it will no longer opperate. Once there is not enough pressure in tank A to opperate the ram,then the valve between tank A and B is opened so as equilisation can take place-->this is the end of the process,we now have 18.2psi in each tank,with a combined volume of 30ltr's.

[LibreEnergia]

Absolute rubbish.
The work done to raise the pressure in tank A IS the stored energy in tank A. The efficiency of the compressor to perform this work has nothing at all to do with the efficiency of the DUT.

Actually it have everything to do with 'it' unless you are going to simply ignore the fact that to operate a full cycle you MUST restore all the potential in the tanks back to the starting condition.

The compression part of the cycle has to be extremely efficient since the only net energy you have available is that,that was able to enter the system from the environment during the expansion phase.  Unfortunately CARNOT rules here. That amount of 'ingested energy' will never be sufficient to perform the work required to recompress the system.

If you are simply going to ignore that requirement, then it is not even worth continuing to analyse. There has never been a dispute that you could extract work out of some store of potential without bothering to replenish it.