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Author Topic: Rosch taking orders on OU Bouyancy device.  (Read 210203 times)

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #330 on: May 02, 2015, 08:07:39 AM »

Can some one tell me what the amount of energy in joules is required to compress a vessel volume of 1 ltr to 290 psi gauge pressure?-just ambiant air as gas.
Just wanting to see how that compares to a bouyant systems graph i have drawn up.

The answer depends on how you compress it and the properties of the gas being compressed.

For air, isothermal compression uses less energy compared with adiabatic, but takes an infinite amount of time vs adiabatic that can occur rapidly but requires perfect insulation.

During compression the work required is equal to the integral of P.dV from V1 to V2

The isothermal case is easiest to  derive.  Ideal gas law tells us P = nRT/V and the integral evaluates as W = -nRT ln(V1/V2)

So, to answer the question and actually put numbers to it you need to tell us how the air is compressed.





tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #331 on: May 02, 2015, 08:14:47 AM »

Quote
That is utter freaking nonsense.  Learn to use a protractor.Or just drink more Kool-Aid.
A protractor to get a 757 to do a 330* spiral dive at 470MPH,and nail the target-->thats a good one MarkE lol. Even TK has stated that the maneuver that was supposedly carried out on the pentagon attack would be extreemly dificult even for an experianced pilot-but the box cutter boys pulled it off no problem.
Oh,and by the way,we dont have kool aid here in OZ-we deal in facts-unlike your self.

 
Quote
Is your bet on nano thermite or dustification?  Maybe you think invisible giant space zombies pounded the buildings down with their giant invisible fists.

You have completely lost the plot MarkE. Your fire scenario is absolute junk-can never happen. Im sorry,but office furniture cant burn hot enough to turn steel to molten metal-like seen flowing out of the window's,and dug up from the rubble weeks later still molten hot.

You have failed to provide a picture(just 1) of another steel framed highrise building that has collapsed from fire-i have provided many that have not-even after days of burning.
In fact,you have failed to deliver on any credible evidence that supports your theories.

I am done debating this with you any further,as you are simply blind to the fact's that may upset the ballance of your perfect little world.

Now,if you would care to get back on topic,i have posted a question on the energy(in joules) required to fill a 1 ltr volume vessel to a pressure of 290psi gauge pressure. This IS in relation to this topic,and i believe this is something you can get right.

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #332 on: May 02, 2015, 08:20:34 AM »
If you are trying to say that all of the input energy is lost to waste heat due to compression and expansion, then that is also wrong.
No it's lost to displacement of the water.  the heat gained/lost results in overall isothermal.  "Since air is a great insulator" ( http://www.physicsclassroom.com/class/thermalP/Lesson-1/Rates-of-Heat-Transfer  ) the heat it gets is going to remain for the few seconds it's compressed... it takes a long time to freeze a bottle of air so it's pressurized at room temp for instance (did this so my water vortex bottles had pressure).

what few milli-watts it sheds to the low water in pulled back when it decompresses at the top... so again tempurature difference is irrelavent.

but still none of that is used to make the motion.

There's no energy in displacement.

tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #333 on: May 02, 2015, 08:26:00 AM »
The answer depends on how you compress it and the properties of the gas being compressed.

For air, isothermal compression uses less energy compared with adiabatic, but takes an infinite amount of time vs adiabatic that can occur rapidly but requires perfect insulation.

During compression the work required is equal to the integral of P.dV from V1 to V2

The isothermal case is easiest to  derive.  Ideal gas law tells us P = nRT/V and the integral evaluates as W = -nRT ln(V1/V2)

So, to answer the question and actually put numbers to it you need to tell us how the air is compressed.
Thank you LE for the civil answer.
I mean to use a standard compressor-a high efficiency one.I know there will be losses in heat,but what is the energy we have in that 1ltr vessel at 290psi gauge pressure. Enviromental temperature will be close to 8*C

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #334 on: May 02, 2015, 08:44:32 AM »

Anyway-back on track here.
Can some one tell me what the amount of energy in joules is required to compress a vessel volume of 1 ltr to 290 psi gauge pressure?-just ambiant air as gas.
Just wanting to see how that compares to a bouyant systems graph i have drawn up.

290psi is 19.7atm.  So 20x the pressure... which is 20L to 1L ...

a standard cubic foot represents 1.19804 moles  ( http://en.wikipedia.org/wiki/Standard_cubic_foot )
1 cubic foot = 28.3168L

n (moles of gas in 20L) =  (0.706 cubic-feet) * 1.19 = 0.84
R  8.3145 J/mol K    ( http://en.wikipedia.org/wiki/Gas_constant )
T 291.3333K (18.3333C)

ln(V2/v1)  (20L/1L) -2.995732274

= 0.84 * 8.3145 * 291.3 * -3
= 2031.491634J

divide by seconds to get watts.
---------------
applying these revisions to my spreadsheet to come out to J and therefore W/s increases COP to 1600 from 400 :) 
Just looks better all the time.

tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #335 on: May 02, 2015, 08:54:47 AM »
290psi is 19.7atm.  So 20x the pressure... which is 20L to 1L ...

a standard cubic foot represents 1.19804 moles  ( http://en.wikipedia.org/wiki/Standard_cubic_foot )
1 cubic foot = 28.3168L

n (moles of gas in 20L) =  (0.706 cubic-feet) * 1.19 = 0.84
R  8.3145 J/mol K    ( http://en.wikipedia.org/wiki/Gas_constant )
T 291.3333K (18.3333C)

ln(V2/v1)  (20L/1L) -2.995732274

= 0.84 * 8.3145 * 291.3 * -3
=
divide by seconds to get watts.

Quote
2031.491634J

Well if that is the correct answer,we can take that 2031.491634 joules,and get it to perform 5780.5 joules of work by using a bouyant device. It will be slightly higher than this,as i never took into account temperature rise as the vessel rises closer to sea level-the surface.

d3x0r

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #336 on: May 02, 2015, 08:58:25 AM »
Well if that is the correct answer,we can take that 2031.491634 joules,and get it to perform 5780.5 joules of work by using a bouyant device. It will be slightly higher than this,as i never took into account temperature rise as the vessel rises closer to sea level-the surface.
temperature falls on decompress.. rises on compression...
(canned air gets cold, decompressing)


tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #337 on: May 02, 2015, 09:21:49 AM »
temperature falls on decompress.. rises on compression...
(canned air gets cold, decompressing)
Yes,but the ocean water temperature rises as you get closer to the surface. At 200 meters deep,the ocean temperature this time of year in the indian ocean is around 8*C,and near the surfact it is around 18*C-that is a 10*C rise in temperature which will be transfered through the vessel to the gas. In my experiments of late,i have seen how fast enviromental temperatures can raise or lower the temperature of a gas in a sealed vessel/tank.

markdansie

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #338 on: May 02, 2015, 09:28:18 AM »
In reply regarding jams Qwok h spent jail time. His devices never worked or panned out, I have covered many articles on him over the years.

Mark

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #339 on: May 02, 2015, 09:41:56 AM »
Thank you LE for the civil answer.
I mean to use a standard compressor-a high efficiency one.I know there will be losses in heat,but what is the energy we have in that 1ltr vessel at 290psi gauge pressure. Enviromental temperature will be close to 8*C

You may want to check the numbers but working in SI,

290 psi gauge is 2100.8 kPa absolute.

during isothermal compression PV = constant so 2100.8 * 1 litre = 101.3 * x => starting volume was 20.73 litres.

The amount of air is n = PV/RT where R = 8.313  J  K-1 mol-1,  and T = 280.15 K (8 Celsius) =   (101.3 * 20.73) / (8.313 * 280.15 )  =  0.9 mol of air.

So,  work performed during isothermal compression = -  nRT ln(V/Vo) = -0.9 * 8.313 * 280.15 * ln(1/20.73) = 6354.2 J or 6.35 kJ

You compressor will be less efficient than that so adjust by whatever efficiency factor you think is appropriate,

lets say it is 65% efficient so 6.38 / .65 = 9.8 kJ

The value is moot though. The same maths that allows you to calculate this number can also be used to deduce that you can never achieve more output than input and the overall efficiency is proportional to the temperatures of the hot and cold sources.

There is energy to be extracted by utilising the difference in temperatures between water at the surface and at depth. Overall efficiency is low though because the temperature difference is low. A very large structure that did not pump air using and external compressor, but rather was the compressor itself would be the best for achieving maximum efficiency.

 


minnie

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #340 on: May 02, 2015, 09:54:34 AM »



    I can remember the day James Kwok "flew into a temper"
     The people that flew into the building were obviously highly competent and had
   planned the whole thing meticulously.
     I have come to dislike anything to do with Peswiki.
   I've lost faith in the Tinman.
      Try putting a 7kw motor on those buckets and water contraption and see what
   happens! My sewage machine runs off a half kw motor and it's amazing how that
  tiny motor turns a 6ft x 12ft rotor that weighs a ton or more.
     It's amazing how much current that little motor uses on no load versus on load.
   There is very little difference,ie. efficiency must be very poor,try telling that to the
  motor-generator crowd.
    Sorry Tinman-I really do appreciate your enthusiasm!!!!!!
                 Elderly farmer John.


tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #341 on: May 02, 2015, 10:27:54 AM »
In reply regarding jams Qwok h spent jail time. His devices never worked or panned out, I have covered many articles on him over the years.

Mark
He spent time in jail for a land rental bungle,not because his device dosnt work. Im sure the UN it self wouldnt promote a non workable device to the world.

tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #342 on: May 02, 2015, 10:30:48 AM »


    I can remember the day James Kwok "flew into a temper"
     The people that flew into the building were obviously highly competent and had
   planned the whole thing meticulously.
     I have come to dislike anything to do with Peswiki.
   I've lost faith in the Tinman.
      Try putting a 7kw motor on those buckets and water contraption and see what
   happens! My sewage machine runs off a half kw motor and it's amazing how that
  tiny motor turns a 6ft x 12ft rotor that weighs a ton or more.
     It's amazing how much current that little motor uses on no load versus on load.
   There is very little difference,ie. efficiency must be very poor,try telling that to the
  motor-generator crowd.
    Sorry Tinman-I really do appreciate your enthusiasm!!!!!!
                 Elderly farmer John.
I in no way support the device being discused in this thread-no evidence to back it up.
I also in no way support peswiki or sterling allan.
I do my own thing,and my own reserch,and as it stands now,all the evidence supports the hidro+ unit.

tinman

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #343 on: May 02, 2015, 10:43:20 AM »
You may want to check the numbers but working in SI,

290 psi gauge is 2100.8 kPa absolute.

during isothermal compression PV = constant so 2100.8 * 1 litre = 101.3 * x => starting volume was 20.73 litres.

The amount of air is n = PV/RT where R = 8.313  J  K-1 mol-1,  and T = 280.15 K (8 Celsius) =   (101.3 * 20.73) / (8.313 * 280.15 )  =  0.9 mol of air.

So,  work performed during isothermal compression = -  nRT ln(V/Vo) = -0.9 * 8.313 * 280.15 * ln(1/20.73) = 6354.2 J or 6.35 kJ

You compressor will be less efficient than that so adjust by whatever efficiency factor you think is appropriate,

lets say it is 65% efficient so 6.38 / .65 = 9.8 kJ

The value is moot though. The same maths that allows you to calculate this number can also be used to deduce that you can never achieve more output than input and worse still the overall efficiency is proportional to the temperatures of the hot and cold sources.

 
So 6.35kJ of energy is stored in our 1 ltr vessel at 290psi.,__im guessing the 6.38 is a typo?.
And 3.45kJ is lost via the compressors efficieny asuuming it is around 65% efficient.

And is it right that to halve the pressure,we double the vessel volume ?

Thanks LE

LibreEnergia

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Re: Rosch taking orders on OU Bouyancy device.
« Reply #344 on: May 02, 2015, 10:53:45 AM »
So 6.35kJ of energy is stored in our 1 ltr vessel at 290psi.,__im guessing the 6.38 is a typo?.
And 3.45kJ is lost via the compressors efficieny asuuming it is around 65% efficient.

And is it right that to halve the pressure,we double the vessel volume ?

Thanks LE

Yes 6.38 should have been 6.35. Halving pressure will double volume.

I see here that the device you seem to be describing is similar to an OTEC generator of which a few exist in Hawaii I believe.

There is nothing to stop you utilising buoyancy to take advantage of the temperature gradient in the ocean to generate energy. However due to the small difference in temperature overall efficiency is low. Unless pumping losses are low it will not function.

The solution is to make the device very large and make it the compressor itself rather than use small external pumps to evacuate a chamber at depth.