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Author Topic: Non Conservation of Energy Research  (Read 960 times)

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See diagram below.

There really is no rocket science going on here.


Brad


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Brad, you said:

Quote
The other thing that you should be looking at,is the !now! 2 half full tanks of compressed gas can do the same amount of work as that of the one full tank of compressed gas--nothing is lost in the process,and so,work was done without loss of energy.

IMHO I think herein lies the misconception. Two  one liter tanks at 20 psi do not have the same energy as a full one liter tank at 40 psi because of the non-linear nature of the air spring.

That a 40 psi tank bled down into another one liter tank leaves 20 psi in each shows that energy was actually lost in the process.

The same goes for capacitors, properly transferring / sharing the energy from one capacitor to the other you wind up with more than half voltage in each, done wastefully and you wind up with around half the voltage.

The energy formula for capacitors is 1/2 CV^2 and it is the square function that is the "gotcha'.

The formula for air tanks probably follows a similar rule with a square function or fudge factor because of the non-linear nature of the air spring / container pair.

You can feel this non-linear effect with a simple bicycle air pump if you close off the end of the hose. It gets progressively more difficult to compress the air in the cylinder. The first couple of inches are easy then it's like hitting a brick wall.

Compressors work very hard trying to get that last little bit of pressure into a tank when topping it up to full pressure. If the air spring were linear, the compressor wouldn't care.  Log the power input to a compressor as it charges a tank to get the full story.

I think temperature of the gas must be considered to get a linear relationship.

Of course I could be wrong about all this, just using my experience and tests as an indicator.

regards

edit: Boyle's original data graph attached

p.s. I think where we need to look is the nature of the momentary temperature excursion when a gas is compressed.
« Last Edit: 2018-06-27, 17:55:59 by ion »


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The process i am showing here is an Isothermal process (as clearly stated in my first post),and in an isothermal process,the internal energy of the expanding system must remain a constant.
In other words,when the temperature is held at a constant,the internal energy remains the same.
As the pressure in tank A decreases,the temperature of the gas decreases,but as the pressure increases in tank B,the temperature of the gas increases.
What is lost in one,is gained in the other-->energy is conserved.
I fully understand that.
Quote
But regardless of that,you seem to be missing the point here.
What i am showing is,there is no change in outcome whether or not that gas dose work when flowing from one tank to the other.
It makes no difference whether or not that air motor/generator/load is there--the outcome is the same.

The other thing that you should be looking at,is the !now! 2 half full tanks of compressed gas can do the same amount of work as that of the one full tank of compressed gas--nothing is lost in the process,and so,work was done without loss of energy.
Without loss of energy according to our known gas laws.  What I have tried to point out is those laws have a limited application in the wider scheme of things.  You seem to think that you can do it again by discharging into twice the new volume, than again and again and so on ad infinitum, and you can't.  Because there is some end point to this, because there is a limit to the amount of energy you can get, I say that the 2 half full tanks cannot do the same amount of work as that of the one full tank of compressed gas--I say something is lost in the process.  You may not be able to measure that loss so you are content to follow the gas laws as they stand. 

Irrespective of our different views on this the other point remains, you have gained some energy but to be usefully OU you must get back to the starting point without giving back that energy.  The gas laws telling you that the energy in the two tanks is the same as the starting energy doesn't solve that problem.  How do you pump back that gas without expending the energy you gained?  You say you have a scheme to do that, it'll be interesting when we see that.

Smudge
   

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 author=ion link=topic=3640.msg68552#msg68552 date=1530108207]

 

Boyle's law states that when temperature is a constant,then P1V1=P2V2

Quote
Compressors work very hard trying to get that last little bit of pressure into a tank when topping it up to full pressure. If the air spring were linear, the compressor wouldn't care.

That is because the gas is being heated as well as being compressed.
The extra energy being used by the compressor,is due to the rise in dissipated heat energy.

As stated in my first post,temperatures remain a constant,and there for the P1V1=P2V2 hold's,and is a linear relationship.

Quote
p.s. I think where we need to look is the nature of the momentary temperature excursion when a gas is compressed.

To quote my first post
Quote: 1- Temperatures will remain a constant,where at the start of each test, the compressed gas temperature in tank A will be allowed to settle to ambient temperature(room temperature)
At the end of each test,the gas temperature in both tanks will once again be allowed to settle to ambient temperature.
This is to eliminate the calculation of dissipated heat energies during each test.


Quote
Log the power input to a compressor as it charges a tank to get the full story.

Will do,and also the temperature of the compressor.

Quote
The formula for air tanks probably follows a similar rule with a square function or fudge factor because of the non-linear nature of the air spring / container pair.

You can feel this non-linear effect with a simple bicycle air pump if you close off the end of the hose. It gets progressively more difficult to compress the air in the cylinder. The first couple of inches are easy then it's like hitting a brick wall.

That is because the temperature is not a constant.



Brad


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Brad,

Is there a way to compress a gas without creating a temperature rise?

Regards,
Pm
   

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Brad,

Is there a way to compress a gas without creating a temperature rise?

Regards,
Pm

Yes,there are a few ways.
You compress it slowly,so as the generated heat can be dissipated away through the vessel.
You can pass the compressed gas through an intercooler/heat exchanger before it enters the vessel.
You draw the gas through an intercooler/heat exchanger before it enters the compressor
And there is also  Adiabatic cooling,and this method is very efficient.

At work,most of our power tools are air power tools,such as rattle guns.
We pump 110psi into those thing's,for minutes at a time.
These rattle guns have vain type motors in them,where cylinder pressures are at 110psi--and yet the guns freeze up if used for to long a duration.
This means that the Adiabatic cooling effect at the exhaust is far greater than the pressure heating within the cylinder.

But to avoid all this extra equipment,and to make the DTU very easy to build and take energy level measurements,i stated in my very first post that during each cycle of the test,all gasses would be allowed to cool or rise to ambient temperature.

The only way you can loose energy in a tank to tank transfer,is by way of radiated heat energy.
Allowing the temperature to remain a constant at each test point during a complete test cycle,we eliminate that loss,and so the energy value remains the same.


Brad


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