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#34090 - 04/26/10 03:08 PM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Maybe the information on this page:

http://www.lhup.edu/~dsimanek/museum/themes/buoyant.htm

is useful to see why a buoyance-driven-perpetuummobile doesn't work.

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#34091 - 04/26/10 03:16 PM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
the greatest amount of vacume that I can get using the hand pump
is 27 inches / -90 kPa , and at this vacume pressure the sides or surface of the flask is mostly covered by bubbles at this pressure.

if I agitate the flask then the larger bubbles float up and go out of solution , but new tiny bubbles form.

however if I strike the bottom of the flask against the bottom of a glass that the flask is sitting in , then thousands of tiny
bubbles stream up to the top !

if I continue striking the flask I keep getting the streams of thousands of tiny bubbles.

but the vacume pressure drops from -90 kPa to -80 kPa

at -80 kPa no more bubbles form even if I strike the flask.

ok I looked it up

my guage also reads kpa which at 30 inch guage
it is at -100 kPa guage.

the first stroke of the pump creates a -62.5 kPa vacume.
the second stroke of the pump creates a -82.5 kPa vacume.
the third stroke of the pump creates a -90 kPa vacume.

90 kPa is the highest vacume it can accomplish.

atmospheric pressure is 101.3 kPa

however 30 inch / -100 kpa can be exceded using a rotary vane pump and for higher vacume pressure a staged vacume pump
can be used.

.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34092 - 04/26/10 03:31 PM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
paul Offline
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the applications on that page do not require energy input
ie .. there is no force applied.

we are discussing proven methods here.

if you can use energy to displace water at the bottom
wthout exposing that energy to the water pressures.

then you can recover that energy as the buoyant force rises up through the water.

-1 + 1 = 0

now if the buoyancy is due to hydrogen gasses at the top of the water then you still have the energy in the hydrogen gasses to burn.

=1
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34095 - 04/27/10 05:31 AM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Yea me and Paul are talking about something with an extra level of complexity, so it's harder to analyze than those examples.

The big difference is when the bouyant 'buckets' get to the top, they don't need energy to reduce their density and make them sink again, in fact we can extract energy while doing this by burning the gas and converting it to liquid water.

I suspect the problem is the energy required to convert this liquid water back into gas at the bottom, but havn't quite seen how that can use so much energy yet.

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#34096 - 04/27/10 05:33 AM Re: Harnessing the power of Hydrogen + Gravity [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
That's a pretty cool vacuum pump. I don't think it's necessary to worry about going beyond -90kPa. That leaves only 10% of the air in the chamber. Pretty close to none.

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#34099 - 04/27/10 02:44 PM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Instead of using a complex electro-chemical reaction to convert water to some gas, I would like to see an analysis of a simple mechanical equivalent.

For example: consider a box with a volume of 3 units (cm, m, whatever). Inside this Box, on each end, are two smaller boxes, each with a volume of 1 unit.

Both this smaller boxes are connected via a spring.

The weight of the object could be 1 unit (1g, 1kg, 1 metric ton, whatever)

By use of energy we are able to extend the inner boxes. This way we can reduce the density of this object from 1/3 to 1/5.
The energy to reduce the density is stored in the spring.

In my opinion this mechanical object features the same basic attributes as you are using from the H2O-molecules.


Lets assume this object is submerged in an environment with a density of 1/4 (in g/m or kg/m).

In its extended state it will float up.
Since it displaces 5 units of volume, each with a weight of 1/4 It should experience an uplifting force equivalent to 5*1/4 - 1 = 1/4 units of weight.
Thats the weight difference between the displaced surrounding and the weight of our object.

On the way down our object will be in its retracted state.
Now the downward force is equivalent to 3*1/4 - 1 = -1/4 units of weight.

Lets assume our apparatus has a hight of 100 units of length (cm, m, whatever).

The energy gained on the way up should be force * distance = 1/4 * 100 = 25 units of energy.
On the way down we have a weight of 1/4 units.
E = m * h --> 1/4 units of weight * 100 units of distance = 25 units of energy.


So we have gained 50 units of energy, haven't we?

On the other hand: our object is ALWAYS submerged in the environment of 1/4 density.
How much energy do we need to switch our object from the retracted to the extended state?

As our apparatur is 100 units of length in height, the surrounding pressure on the bottom should be: p = rho * h.
p = 1/4 * 100 = 25 units of force.

To extend ONE of the two inner boxes we would have to spend 25 units of energy, to move the inner box outward against the surrounding pressure.
We are doing this for both boxes: voila 50 units of energy are used.

----

By using an electro-chemical reaction to change the density of your objects (H2O-molecules) you are obfuscating the need to use energy to change the density.

I guess you could complicate things furthermore by adding areas of different pressure.

------

The conservation of energy holds true for all experiments so far. That's why its called a law of nature - not because of some conspiracy.

Even if we observe some strange process where energy seems to vanish or appear: it is most likely we just don't understand the process.
For example: The neutrino was predicted because of an apparent loss of energy during radioactive decay.


Edited by Momos (04/27/10 02:51 PM)

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#34100 - 04/27/10 08:08 PM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
paul Offline
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Registered: 03/21/06
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Momos

the difference between the spring example you have and what we are zeroing in on is the extra energy that can be had durring the process.

ie.... the HHO

we are paying for the energy we use as we go , and generating extra energy durring the process.
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34101 - 04/27/10 08:18 PM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
paul Offline
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Registered: 03/21/06
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Kallog

I've got it all figured out now but Im making some drawings so I will post them up when Im finished.

I have reverted back to buoyancy and kept the gravity also.

and the design is not obstructed by water pressure durring the HHO generation.

----------------
I have been trying to figure out something that seems odd to me.

when I pump up the vacume in the flask containing water the bubbles form , and the vacume bleeds down because of the bubbles forming.

so I pump it back up , it bleeds back down again because of bubbles forming but fewer and fewer , eventually when I pump up the vacume there are no more bubbles that form.

so something is happening to the water.

I just cant figure it out...

got any ideas?
_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34102 - 04/27/10 09:55 PM Re: Harnessing the power of Hydrogen + Gravity [Re: paul]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Originally Posted By: paul

the difference between the spring example you have and what we are zeroing in on is the extra energy that can be had durring the process. ie.... the HHO
we are paying for the energy we use as we go , and generating extra energy durring the process.


It is the same principle.
In my mechanical equivalent you insert energy to extend the spring. The contracting spring can be used at any time to get SOME of this energy back.

That's the same as your idea of using energy for electrolysis and regaining this energy at a later time.

In both cases the problem persists: reducing the density of anything (Molecules, inflatable balloons, mechanical devices) in some medium will need exactly as much energy as you are gaining by letting it lift by buoyancy and fall down again.

My example uses just a mechanical device to see more clearly why it takes so much energy.

Just keep in mind: you will need more energy for splitting water to HHO then you can regain by recombination of HHO to water. (At least as long as the splitting is done under higher surrounding pressure then the recombination).

----

You water:
Couldn't this be dissolved gases, which are released in a low pressure environment? Sounds to me like the stuff happening to divers when they inhale gases at high pressure and then surface to lower pressure.



Edited by Momos (04/27/10 10:01 PM)

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#34104 - 04/28/10 02:18 AM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
paul Offline
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Registered: 03/21/06
Posts: 4136
Momos
you compress the spring at the top and use 1 unit if energy
and the box sinks.
then if you decompress the spring under water so that it will expand and float up , then you are making the water level at the top increase , and you are lifting the water using the 1 unit of energy you put in compressing the spring.

you still have the energy from buoyancy as it rises to the top.

so you can compress the spring again.

but your problem is that you are working against the water pressures at the bottom.

if you didnt have to expend the 1 unit of energy at the bottom pushing against the water pressure
you could compress the spring at the top again.

and if you let the box rise more you could get more energy out from buoyancy.

this would be your extra energy over 100% efficicency.












_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34105 - 04/28/10 04:19 AM Re: Harnessing the power of Hydrogen + Gravity [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Yea that'll just be water boiling into water vapour becuase the boiling point drops with low pressure. Then you have to suck that vapour out, bringing it back to low pressure, and allowing more to boil.

Maybe this is a serious issue, or maybe you can just do electrolysis on the water vapour?? Seems unlikely because it won't be very conductive.

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#34106 - 04/28/10 04:39 AM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Yes that agrees with what I was imagining. But how can you translate it into H and O molecules?

We're aware that when the 'box' is at the bottom, and you're reducing its density to make it float:
X joules of electrical energy is used, which goes towards
Y joules of chemical energy and
Z joules of work lifting the column of fluid above it.

Does X = Y + Z + losses? Conservation of energy says yes, but we're not allowed to rely on that. If we rely on that law then the whole system is automatically, and uninterestingly unworkable without any analysis.

From Paul's calculations which I checked too, it doesn't appear to add up right.

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#34107 - 04/28/10 04:46 AM Re: Harnessing the power of Hydrogen + Gravity [Re: paul]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: paul
Momos
and if you let the box rise more you could get more energy out from buoyancy.


Not in Momos's system because it's already rising to the top of the column of water. If he'd had it rise half way, then you'd only need half as much energy to compress the spring at the top of its cycle. But also only get half as much energy out of the generator. So I think in Momos's system it doesn't matter how high you let it float, or how high the tank of water is. As long as the former is no more than the latter of course :P

I suspect this is the same with the HHO, but still can't see how the numbers add up.

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#34111 - 04/28/10 10:33 AM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Originally Posted By: kallog

I suspect this is the same with the HHO, but still can't see how the numbers add up.


Thats why I advise to concentrate on a full circle of one molecule of water (or its mechanical equivalent).
A guess paul wrote somewhere "the volume of HHO is 1800 times larger then water".
Thats where the energy is lost: you have to spend energy to do this expansion against the surrounding pressure.
Electrolysis under high pressure takes more energy then under low pressure.



Originally Posted By: kallog

So I think in Momos's system it doesn't matter how high you let it float, or how high the tank of water is. As long as the former is no more than the latter of course :P


Correct: the higher the tank: the more energy you gain from floating and falling. But the amount of energy necessary to change the density (do the electrolysis) increases accordingly.
Both processes are directly depending on the density of the surrounding and the force of gravity.

Originally Posted By: kallog

From Paul's calculations which I checked too, it doesn't appear to add up right.


Well, I'm unable to calculate the amount of energy (Z)to seperate the atoms of the molecule against the surrounding pressure. Thats why I used my mechanical boxes as an equivalent. Do my figures add up?


Edited by Momos (04/28/10 10:37 AM)

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#34123 - 04/28/10 12:17 PM Re: Harnessing the power of Hydrogen + Gravity [Re: paul]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Sorry, probably my english is not good enough to understand everything. But let me try:

Originally Posted By: paul

you compress the spring at the top and use 1 unit if energy
and the box sinks.
then if you decompress the spring under water so that it will expand and float up ,


Thats not my initial model.
In my model I use the energy at the bottom to stretch the spring. And at the top I get some of this energy back by letting it recontract.
Same as you using energy for electrolysis at the bottom and regaining some of this energy at the top.


But, ok. Lets assume we are compressing the spring at the top, using 1 Unit of energy. The Box will now fall downwards. At the Bottom we have the let the spring stretch again.
The problem is: at the top we have less surrounding pressure then at the bottom. So the energy used at the top to compress the spring is NOT ENOUGH to let the spring stretch on the Bottom.

Lets quantify this: We build an offshore-box-driven-power-plant in some part where the oceans depth is 10km.


Tower height: 10.000 m.
Density of Water: 1000 kg / m

Weight of Box: 6000 kg.
Volume of Box at the top: 3 m
Density of Box at the top: 2000 kg/m (will sink)
downward force while sinking: [6t - 3m * 1t/m ] * 9.81m/s = 29.43 t*m/s ~ 30 kN

Volume of Box at the bottom: 12 m
Density of Box at the bottom: 500 kg/m (will float up)
upward force while floating up: [12m * 1t/m - 6t] * 9.81m/s = 58.86 t*m/s ~ 60 kN.

Energy gained by floating up and falling down:
10.000m * 60kN + 10.000m * 30kN
= 900.000.000 N*m = 900 MJ.
(or 882.9 MJ using the exact numbers)


How much energy do we need to convert the box from the sinking to the floating state?

At the bottom we have to extend the volume of our box from 3m to 12m. So we have to push 9m of water away.
The surrounding pressure at the bottom would be: 10.000 m * 1000 kg/m * 9.81 m/s = 9.81 * 10^7 N/m = 9.81*10^7 Pa = 981 bar.

The energy needed to gain 9m of volume at this pressure would be: 9m * 9.81*10^7 N/m = 88.29*10^7 N*m = 882.9 MJ.


All generated Energy is lost if you try to complete a full cycle.



If we try your latest approach and compress the spring at the top we have the same problem.
Either we use a light spring, easy to compress.
But then the stored energy in the spring won't be able to overcome the surrounding pressure. The box won't float up again. Or we use a heavy spring, but then we have to apply the needed energy at the beginning for compressing it.






Originally Posted By: paul

then you are making the water level at the top increase , and you are lifting the water using the 1 unit of energy you put in compressing the spring.
[...]
and if you let the box rise more you could get more energy out from buoyancy.


Now, this is a different model!
Now you are complicating things by changing the water level during the floating phase.
I guess now the problem is:
a) inserting the box at the bottom of the water-tower at the start of the second run. (pushing against the now elevated water level means pushing against more pressure loosing all energy gained by increasing the falling and rising distance.
b) You could try to sink the water level again, before inserting the box at the bottom. But now you have the problem of re-elevating the water level. You are loosing again exactly the same amount of energy.
c) You could use water from a separate source (like rain), but then you have just an open system, gaining nothing more then the potential energy stored in raindrops.

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#34124 - 04/28/10 12:43 PM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Hi Momos. I think it's getting confusing with all the different systems. I know you won't want to read all the history, but me and Paul have discussed various configurations, including doing it underwater, and come to one he described clearly in message #34053.

I noted that it required more height than the atmosphere had. But this is where my thoughts are at this stage:

a) Convert a liter of water to HHO, let it float up (throw away that energy), burn it back to water (throw away that energy), then drop the water back to the ground (recover that energy).

b) We know how much energy you can get from a liter of water falling through a height of 978 miles (4212 W-hr)

c) We know how much energy it takes to convert a liter of water into HHO at atmospheric pressure (4212 W-hr)

d) If you had an atmosphere that was >978 miles high, but had a pressure at ground level the same as our atmosphere, then it looks like you come out better than even. This atmosphere would have to be made of a gas less dense than air, and a tiny bit denser than HHO.

e) You don't even need such an extreme because you can recover more from the "throw away that energy" stages in a).

f) There's another option of pumping out a vacuum to generate the HHO in.

Maybe my requirement in d) is impossible. I have to work that out, but I'd rather wait for Paul to present a new complete system.


A big stumbling block is not knowing how much extra energy is needed to produce HHO at higher pressures.

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#34126 - 04/28/10 01:40 PM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
Momos Offline
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Registered: 01/15/10
Posts: 48
Loc: Germany
Thanks for your explanations! smile

> A big stumbling block is not knowing how much extra energy is
> needed to produce HHO at higher pressures.

Thats the point: you need at least as much energy as needed to overcome to surrounding pressure and to enhance the volume.

Thats the whole point of my mechanical box: to make it easier to calculate this needed energy.

The surrounding pressure at the bottom, the buoyance and the falling distance are directly related to each other.



> d) If you had an atmosphere that was >978 miles high, but had
> a pressure at ground level the same as our atmosphere, then it
> looks like you come out better than even.
> This atmosphere would have to be made of a gas less dense than
> air, and a tiny bit denser than HH

The pressure at groundlevel is directly depending on the density of the atmosphere and the gravitional pull.

pressure = height * density_ath * g
energy_gain = energy_gain_up + energy_gain_down

energy_gain_down = height * g * volume_sinking * density_ath

energy_gain_up = height * g * volume_floating * density_ath

energy_loss = pressure * (volume_floating - volume_sinking)

total_energy = energy_gain - energy_loss

As you can see the total amount of energy can be calculated by this simple formula which is only referring to the basic parameters of your system:

= (height * g * volume_sinking * density_ath) +
(height * g * volume_floating * density_ath) -
(height * density_ath * g * [volume_floating - volume_sinking])

since "height * g * density_ath" is always the same constant factor k, we can write:

= ( k * volume_sinking ) + ( k * volume_floating ) - ( k * [volume_floating - volume_sinking])

I hope it is clear this formula will always result in 0 !
It doesn't matter how big or small this factor of k is!

You can cange the density of the atmosphere, the gravitational pull and the height of your tower in ANY COMBINATION, it will only result in some different constant factor "k".

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#34130 - 04/28/10 11:03 PM Re: Harnessing the power of Hydrogen + Gravity [Re: Momos]
kallog Offline
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Registered: 03/17/10
Posts: 1100
Originally Posted By: Momos
Thats the point: you need at least as much energy as needed to overcome to surrounding pressure and to enhance the volume.

We only know that because we assume conservation of energy. We're not allowed to make that assumption, because then the argument is circular - "the machine won't violate conservation of energy because you can't violate conservation of energy".

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#34131 - 04/29/10 12:39 AM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
paul Offline
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Registered: 03/21/06
Posts: 4136
Kallog and Momos

I have finished and would like to ask you one question.

if I have a glass that is half filled with water.
and I put a straw in the glass , then put my finger over the top of the straw and lift the straw almost all the way out of the water.

1) would the water pressure at the bottom of the glass increase?

although the water height has increased , the water pressure at the bottom of the glass has not increased because the water in the straw is supported by the vacuum in the straw , but the water height from the bottom of the glass to the top of the water in the straw has increased.





_________________________
3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.

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#34135 - 04/29/10 08:43 AM Re: Harnessing the power of Hydrogen + Gravity [Re: kallog]
Momos Offline
Member

Registered: 01/15/10
Posts: 48
Loc: Germany
Originally Posted By: kallog

We only know that because we assume conservation of energy. We're not allowed to make that assumption, because then the argument is circular


But I'm not assuming conservation of energy.
I just calculated the amount of energy needed to extend the box/molecule at the bottom.
It turns out it's the same amount of energy we gained before.
I'm not using some abstract principle but a simple calculation of the necessary forces.

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