Open Systems

[MarkE]

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

Energy isn't lost to entropy.  Entropy saps the ability to convert energy to useful work.

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

If you understand heat pumps, then you should understand the similarities here.

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

The normal way that we do honest accounting is to perform an energy balance through one or more complete cycles.  Since we have not done that for your scheme at best you simply do not know where you are going to end up when an entire cycle is accounted.  As it stands, you have greatly reduced the apparatus' capacity to perform useful work but you haven't done any useful work yet.  As a Gedanken consider what it will take to end up with 10 l of 40psi gauge air at room temperature and how you will get there from where you are.  You have 10 l of 16.2psi gauge, so roughly 30.9 psi absolute.  You need to increase that gas quantity to 40 psi gauge or 54.7 psi absolute.  You can either use both the 20l and 10l tanks to compress atmosphere into a 10l tank, or only the 20l tank contents to do work that will compress atmospheric gas into the 10l tank.

You will need to apply pressure at 40psi gauge + at the end of your recompression.  How man psi gauge do you plan to have at that point?  For sake of argument, say it is 5.0.  You will need an 8:1 diaphragm or piston to develop the 40psi gauge from 5psi gauge.  You surely don't have 8:1 volume to spare, so scratch that.  Try 15.0 psi gauge.  Now you need 8:3.  But you can only afford to use about 6% of your gas, and you are again cooked.  You could try and come up with a clever way to build a variable gain compressor so that just as you run out of gas you get back to 40 psi guage and hope with that you can get enough gas back into tank A.  It won't happen.

 

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?.

While you object that I contend you suffer a mental block, entropy really seems to have you stumped.  A bathtub full of room temperature water is useless towards driving a heat engine that only has other room temperature heat sources / sinks to work against.  That bathtub has plenty of energy in it.

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 have no interest in playing a game of GIGO.  Calculations made based on grossly incomplete or incorrect assumptions yield silly results as we have already just seen.

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.

See above.  I am not interested in GIGO.  You may not consider that information you omit is significant.  You just demonstrated that such details can be very significant.

[tinman]

    You have 10 l of 16.2psi gauge, so roughly 30.9 psi absolute.  You need to increase that gas quantity to 40 psi gauge or 54.7 psi absolute.  You can either use both the 20l and 10l tanks to compress atmosphere into a 10l tank, or only the 20l tank contents to compress do work that will compress atmospheric gas into the 10l tank.While you object that I contend you suffer a mental block, entropy really seems to have you stumped.  A bathtub full of room temperature water is useless towards driving a heat engine that only has other room temperature heat sources / sinks to work against.  That bathtub has plenty of energy in it..See above.  I am not interested in GIGO.  You may not consider that information you omit is significant.  You just demonstrated that such details can be very significant.

The normal way that we do honest accounting is to perform an energy balance through one or more complete cycles.

In order for me to be able to complete the unit to perform a complete number,i need to know where i stand(how much energy i have)after the first half of the cycle-the half that gives us the energy to replenish the 10l tank to it's starting capacity. Without these numbers i cannot calculate the size or dimentions of the cylinder needed to recompress the gas into tank A.

Since we have not done that for your scheme at best you simply do not know where you are going to end up when an entire cycle is accounted.

This is correct,we do not have an answer due to you insisting on needing to know how the entire system opperates. You dont need to know this,i am asking only that you calculate the start and end energy amount avaliable in the two tanks. Knowing how the system opperates has nothing to do with the end result's. Im asking for end result's-not an evaluation on my device.

As it stands, you have greatly reduced the apparatus' capacity to perform useful work but you haven't done any useful work yet.  As a Gedanken consider what it will take to end up with 10 l of 40psi gauge air at room temperature and how you will get there from where you are.

That is what im trying to do-work out how and what i need to get there from where i am ATM. However,i dont know where i am,as you are not interested in calculating the results i gave you for the first half of the cycle.

I have no interest in playing a game of GIGO.  Calculations made based on grossly incomplete or incorrect assumptions yield silly results as we have already just seen

There are no silly game's other than the one you play Mark. Needing to know how the system work's has nothing to do with calculating the energy within the two tank's from the numbers i gave you-thats all i ask. Im not asking you to verify my device,nor am i after your opinion on how my device work's-->just calculating the start energy in tanks A&B,and the end results i have obtained in tanks A&B-thats all you have to do. But i see that this is going to be a never ending game of-->you show me first. Well it looks like it's game over here,and i must say that i have seen better styles of trying to get a glimps of a device and how it work's than the style you just used here. Im sorry,but i dont make deals,nor do i give in to demand's.

Thanks for your time.

[MarkE]

In order for me to be able to complete the unit to perform a complete number,i need to know where i stand(how much energy i have)after the first half of the cycle-the half that gives us the energy to replenish the 10l tank to it's starting capacity. Without these numbers i cannot calculate the size or dimentions of the cylinder needed to recompress the gas into tank A.

You are asking for piecemeal calculations against a model that is not in view.  If you want to go calculate basic quantities on your own try the EngineeringToolBox.

This is correct,we do not have an answer due to you insisting on needing to know how the entire system opperates. You dont need to know this,i am asking only that you calculate the start and end energy amount avaliable in the two tanks. Knowing how the system opperates has nothing to do with the end result's. Im asking for end result's-not an evaluation on my device.

Excuse me but this is your project not mine.  I did not entice you into trying something by making a promise that I have since broken.

That is what im trying to do-work out how and what i need to get there from where i am ATM. However,i dont know where i am,as you are not interested in calculating the results i gave you for the first half of the cycle.

There are no silly game's other than the one you play Mark. Needing to know how the system work's has nothing to do with calculating the energy within the two tank's from the numbers i gave you-thats all i ask. Im not asking you to verify my device,nor am i after your opinion on how my device work's-->just calculating the start energy in tanks A&B,and the end results i have obtained in tanks A&B-thats all you have to do. But i see that this is going to be a never ending game of-->you show me first. Well it looks like it's game over here,and i must say that i have seen better styles of trying to get a glimps of a device and how it work's than the style you just used here. Im sorry,but i dont make deals,nor do i give in to demand's.

Thanks for your time.

I did not lead you into this forest.  LE did not lead you into this forest.  Both of us have explained to you what's further down the road waiting for you.  You have very confidently expressed that something entirely different lies ahead.  Enjoy the journey.

[tinman]

For those that are interested,here i show the venturi effect doubling the output of a small DC generator. This shows the venturi adding energy to the system,and doing work.

https://www.youtube.com/watch?v=xsDXKN4Mosw

[MarkE]

The demonstraton shows that more power transfers to the fan with the valve opened rather than closed.  It does not determine that the power in the outflow exceeds the power consumed generating the inflow. 

If I connect a DC power source to a fixed resistor and that to a variable resistor I can readily show that  a value of the variable resistor exists that maximizes the power transferred to that variable resistor.  What I cannot do is show that the power exceeds the power drawn from the power source.  We would also see that the less power that the load draws from the power source the greater the percentage of power through the variable load resistor is of the diminishing power drawn.