Author Topic: Open Systems (Read 152051 times)

MarkE · « **Reply #345 on:** April 24, 2015, 05:40:53 AM »

Quote from: tinman on April 24, 2015, 03:40:50 AM

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.

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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.

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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 · « **Reply #346 on:** April 24, 2015, 09:28:56 AM »

Quote from: MarkE on April 24, 2015, 05:40:53 AM

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.

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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.

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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.

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.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.

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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 · « **Reply #347 on:** April 24, 2015, 09:53:57 AM »

Quote from: tinman on April 24, 2015, 09:28:56 AM

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.

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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.

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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 · « **Reply #348 on:** April 24, 2015, 11:34:24 AM »

Quote from: MarkE on April 24, 2015, 09:53:57 AM

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

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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.

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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.

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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.

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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?.

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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 · « **Reply #349 on:** April 24, 2015, 12:56:51 PM »

Quote from: tinman on April 24, 2015, 03:40:50 AM

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.

MarkE · « **Reply #350 on:** April 24, 2015, 01:03:27 PM »

Quote from: tinman on April 24, 2015, 11:34:24 AM

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.

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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.

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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.

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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.

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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.

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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 · « **Reply #351 on:** April 24, 2015, 01:37:24 PM »

Quote from: MarkE on April 24, 2015, 01:03:27 PM

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.

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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.

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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.

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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.

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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 · « **Reply #352 on:** April 24, 2015, 02:28:37 PM »

Quote from: tinman on April 24, 2015, 01:37:24 PM

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.

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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.

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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 · « **Reply #353 on:** April 25, 2015, 09:30:18 AM »

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 · « **Reply #354 on:** April 25, 2015, 11:18:50 AM »

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.

tinman · « **Reply #355 on:** April 25, 2015, 12:01:21 PM »

Quote from: MarkE on April 25, 2015, 11:18:50 AM

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.

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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

It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.

LibreEnergia · « **Reply #356 on:** April 25, 2015, 12:27:32 PM »

Quote from: tinman on April 25, 2015, 12:01:21 PM

It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.

This demonstrates nothing other than a more efficient nozzle. Try the same experiment without a venturi but simply vary the nozzle shape and diameter. You'll find exactly the same effect occurring. A nozzle that can generate a higher velocity flow will produce more lift on the fan and hence more output to the generator.

Since you are not actually measuring the energy potential of the source you can't make any pronouncement at all that an energy gain has occurred.

MarkE · « **Reply #357 on:** April 25, 2015, 12:43:09 PM »

Quote from: tinman on April 25, 2015, 12:01:21 PM

It shows exactly that. More force is exerted on the fan blades when the valve is open,thus more work is being done. The power to create the inflow is exactly equal to the power of the out flow with the valve closed->energy can be neither created nor destroyed. The rest of the power consumed to create the inflow is disipated as heat. Shall we go back to the tanks Mark?. I can set this up with the 20ltr tank,and show a run from the tank with and without the venturi open. We could put say 40psi into the tank,and wait for the gas temperature to rest at ambiant temperature,then see how long and how high the power output from our little generator go's with the venturi closed off. We could then do the exact same run with the venturi open. Which do you think will be able to do more work?.

Tinman there is no measure of the energy required to generate the flow in the tube. So that is a dead stop to determining outflow energy over inflow energy right there. It shows only what you see, more power transferred to the blades in one condition rather than the other. We can make the blade spin faster by increasing the power in the tube and/or by improving the impedance match between the blade and the outflow from the tube. Surely you agree that if we were to for example change the angle of the blades that we could change the fan speed a lot without hardly affecting the flow through the tube at all. We can turn that right around by changing the geometry of the nozzle.

tinman · « **Reply #358 on:** April 25, 2015, 01:10:52 PM »

Quote from: MarkE on April 25, 2015, 12:43:09 PM

Tinman there is no measure of the energy required to generate the flow in the tube. So that is a dead stop to determining outflow energy over inflow energy right there. It shows only what you see, more power transferred to the blades in one condition rather than the other. We can make the blade spin faster by increasing the power in the tube and/or by improving the impedance match between the blade and the outflow from the tube. Surely you agree that if we were to for example change the angle of the blades that we could change the fan speed a lot without hardly affecting the flow through the tube at all. We can turn that right around by changing the geometry of the nozzle.

We could indeed change the angle of the blades to make it more efficient,but once again,the venturi open would still produce more torque from the fan. We have a higher gas flow with the venturi open than we do with it closed. Maybe we show the difference where flow angle or geometry dosnt matter. Maybe we show usful work being done by way of the flow of gas lifting a mass in a given amount of time?. It takes more energy to accelerate a given mass in a shorter period of time-dose it not?. So how about we do something really simple,like put a pingpong ball(our mass),into a tube of a given length,and see how fast it travels through that tube with and without the venturi open?.

MarkE · « **Reply #359 on:** April 25, 2015, 01:49:20 PM »

Quote from: tinman on April 25, 2015, 01:10:52 PM

We could indeed change the angle of the blades to make it more efficient,but once again,the venturi open would still produce more torque from the fan. We have a higher gas flow with the venturi open than we do with it closed. Maybe we show the difference where flow angle or geometry dosnt matter. Maybe we show usful work being done by way of the flow of gas lifting a mass in a given amount of time?. It takes more energy to accelerate a given mass in a shorter period of time-dose it not?. So how about we do something really simple,like put a pingpong ball(our mass),into a tube of a given length,and see how fast it travels through that tube with and without the venturi open?.

Changing a variable and noting that the fan speeds only tells us that more power was imparted to the fan blades. Since that can occur for several reasons, it does not tellwhich of those reasons or combinations of those reasons.

Isn't your hypothesis that you can perfom more work than energy you input by plumbing your apparatus to the local environment? If that is what you are trying to get to, then the experiment design effort should try and falsify that as directly and simply as possible.

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