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paul Offline OP
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Me and Kallog have been discussing using the buoyancy
of HHO to recover the electricity that was used in converting water into HHO in the thread
"Harnessing the power of the sun " since the thread has
moved so far off course I have decided to move it to its own thread.

Kallog, since we were having such a hard time with the buoyancy I though I would make an image of a way to use gravity instead , perhaps we can agree on this one and dispose of all the complicated buoyancy and pressure stuff and get on with it.

in the below image the water is converted into HHO at the bottom and then the HHO rises to the top where it is converted back into water as it is used as a fuel.
the water then is put into the containers that were emptied at the bottom and gravity pulls them down.

click the below image for a larger one.
then click the larger one for an even larger one.



since it is the amount of weight pressing downwards that will determine the amount of electricity you can recover
from a HHO generating process , and since the height of this type of system does not require more electricity for a higher system to produce the HHO at the bottom.

and since the HHO is lighter than air and will rise to the top itself without requireing additional electricity
the amount of electricity you can recover or produce or both only depends on how high you want to build it.

or how much FREE ENERGY you can handle.

BTW

you can do the above alot cheaper by using a tall pipe , and let the water flow into the pipe at the top.

then at the bottom you can use a hydraulic motor , just let the pressurized water from the bottom of the pipe flow through the hydraulic motor at the bottom to turn a gear box (at whatever rpm your generator system requires) and then the generator.







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Hi Paul. Good idea about a new thread.

Is the system sealed? If so then the HHO can't float up because there's no denser fluid (air) floating down to displace it.

If it's open to the air, and air sinks down the pipe to push the HHO up, then you have a higher pressure at the bottom than the top. The higher the tower, the more pressure, so the more energy needed to create the HHO. You told me yourself that HHO generation requires more energy at higher pressures.

I think we'll be running in circles until we can quantify it.

Or just build one! Look how cheap and simple it is!! Some bits of wire in a pipe, a water wheel, an old lawnmower motor and an alternator from the car wreckers.

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paul Offline OP
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Kallog
Quote:
Is the system sealed? If so then the HHO can't float up because there's no denser fluid (air) floating down to displace it.


yes the system is sealed between the point where the HHO is generated and the point where the HHO is used as a fuel.

I forgot to include this in the image but the larger pipe
(B2) has air inside it , thats why its larger.

the HHO rises through the air due to its density.
air cannot flow through the pipe (B1) because of the
pressure inside (A1).

and because HHO is lighter than air the HHO will gather at the top of (B2).

the HHO that gathers at the top in (B2) also acts as a
storage area where more HHO can be drawn off when needed
for power surges in the power plant.

depending on the type of power plant you are using
ie.. if using a ICE engine or turbine then you would mix air with the HHO at the top after the HHO has passed through (B3).

unless you have a ICE or turbine that uses only HHO.

for a fuel cell you would not want a mixture.
but you would need a seperator to seperate the hydrogen from the oxygen.

but thats not really a part of this recovery system as it can vary greatly.

but if you were to need to build one that doesnt use a pipe that uses air to raise the HHO , and need to determine the difference in weight of the two
( air and HHO ) I have provided this below.

According to the CRC Handbook of Chemistry and Physics, the density of dry air at 20 degrees C at 760 mm of mercury (one atmosphere of pressure) is 1.204 milligrams per cubic centimeter.

1 cubic foot = 28,316.8467 cubic centimeters.

So, dry air weighs 34,093.48 mg per cu.ft.

Which is about 1.2 ounces per cu.ft.

if the pipe had a cross sectional area of 1 entire sq inch
this means that 144 feet of pipe could house 1 cu/ft of air or HHO at any given time.

so if the pipe is 144 feet tall then there would be a need to supply a force of 1.2 ounces , well not that much.

as HHO weights less than air.
1 cu inch of air weighs .00004 pounds
1 cu inch of HHO weighs .00002 pounds


1 cu ft of water weighs 62.4 lbs
1728 cu inches in 1 cu ft
62.4 / 1728 = .0361 lbs
1 cu inch of water weighs .0361 lbs

water expands 1800 times when converted into HHO
1 cu inch of water weighs .0361 lbs
so 1 cu inch of HHO weighs .0361 / 1800 = .00002 lbs

there are 12 inches x 144 ft of HHO in the pipe so it
weighs (12 x 144) x .00002 lbs = 0.03456 lbs

WOW

so for every 144 feet in height you would need to supply a
constant force of .03456 lbs in order to get the HHO to go into the pipe at the bottom.

Quote:
I think we'll be running in circles until we can quantify it.


well that has that pretty much quantified wouldnt you say

Quote:
Or just build one! Look how cheap and simple it is!! Some bits of wire in a pipe, a water wheel, an old lawnmower motor and an alternator from the car wreckers.



I think that building something should follow design work
after all you need to know how much electricity your system would use in producing HHO in order to design a proper recovery system to recover that electricity.

then theres the percentage that you want or need to recover
if you only want to recover the 22% lost durring conversion to HHO then the system could be much smaller.

if your providing hydrogen for the space shuttle or commercial air lines then your system would need to be much larger.



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Originally Posted By: paul

so for every 144 feet in height you would need to supply a
constant force of .03456 lbs in order to get the HHO to go into the pipe at the bottom.


Cheers for working that out. Yea it isn't much, but the tiny weight has a downside -

After you burn the entire 0.035lb of HHO, you only get 0.035lb of liquid water. You've applied enough power to lift 0.035lb up 144ft, and at the same time the water wheel generates as much power as you can get from dropping 0.035lb down 144ft. It's the same!!

You'd need a 100% efficient water wheel just to provide enough power to pump that gas up pipe B2. So there's nothing gained at all.

If this doesn't sound right, remember the water wheel will be rotating extremely slowly. Its buckets are filling 1800 times slower than the HHO is going up the pipe.



Quote:

well that has that pretty much quantified wouldnt you say

It's half the story, but you completely omitted the energy generated. It has no value unless it generates more than it consumes.

Quote:

I think that building something should follow design work

So what? Mortgage your house and hire an engineer. No expense is too great. You're forgetting again that after you have it working, you'll be RICHER THAN GOD!! Not only that but you'll have solved all of humanity's biggest problems. Are you too selfish to spend some of your time and money to give billions of people better lives?

Or do you expect it might fail?

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paul Offline OP
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Kallog

Originally Posted By: Kallog
After you burn the entire 0.035lb of HHO, you only get 0.035lb of liquid water. You've applied enough power to lift 0.035lb up 144ft, and at the same time the water wheel generates as much power as you can get from dropping 0.035lb down 144ft. It's the same!!

You'd need a 100% efficient water wheel just to provide enough power to pump that gas up pipe B2. So there's nothing gained at all.


Originally Posted By: paul
yes the system is sealed between the point where the HHO is generated and the point where the HHO is used as a fuel.

I forgot to include this in the image but the larger pipe
(B2) has air inside it , thats why its larger.

the HHO rises through the air due to its density.
air cannot flow through the pipe (B1) because of the
pressure inside (A1).


I did include the calculations to determine the amount of force required to just lift the HHO up the 144 ft distance.

but I noted that this was if you needed to use it to determine
the amounts of force
to do so , not that it should be done.

since the HHO would simply float up the 144 ft distance
and you have agreed that the water falling would supply the energy that would be used up if it were lifted up 144 ft , the only cost in HHO production would be the cost at (A1).

this cost is recovered as the fuel is burned at the top.

and since HHO floats on air due only to its density as you
have admitted to in the previous thread , there is no reason that the system could not be higher than 144 ft.

and the higher it is the more energy you can produce as the water falls.

even if the water falls slowly , the higher it is the more torque the weight of the water will present to a shaft.

and high torque low rpm can be converted to low torque high rpm.

Quote:
hire an engineer


why would I go to all the trouble to just have a incompetent engineer work all the efficientcies out of the system , no I wouldnt hire any engineers.

sorry !

.


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Originally Posted By: paul

I did include the calculations to determine the amount of force required to just lift the HHO up the 144 ft distance.

but I noted that this was if you needed to use it to determine
the amounts of force
to do so , not that it should be done.

OK. So we've established that pushing it up is not an option. That's progress, we've closed off one dead-end avenue.

Quote:

since the HHO would simply float up the 144 ft distance
and you have agreed that the water falling would supply the energy that would be used up if it were lifted up 144 ft , the only cost in HHO production would be the cost at (A1).


You calculated the force per square inch required to hold up a 144ft column of HHO. This is the pressure that the HHO generator needs to work at. If we replace some or all of the HHO with air, its weight is greater, so the HHO generator has to work at a greater pressure. That's even worse. It doesn't matter that the HHO is floating up, we still have to push it into the pipe against the back-pressure of that column of air, which takes the same power as lifting the column of air itself, and in fact is what we are doing.

Don't believe me? Try floating helium into an inflated car tyre. Without a high pressure pump (using energy) it doesn't go in at all, despite being bouyant.


-----

I think an HHO generator would have maximum efficiency operating in a vacuum. You suggested something close to this yourself.

That means any pressure higher than absolute zero requires additional electrical power input. This column of air/HHO must necessarily have a pressure at the bottom of it which increases in proportion to the column's height. Every extra foot of height you give it increases the power consumption of the HHO generator (ignoring the atmosphere, which can be sealed out of the system anyway).

You could achieve equal or better efficiency by placing the HHO generator at the top of the tower. You would then have to remove the water wheel. But that destroys the whole idea.

That argument isn't convincing unless I show that the extra power consumed by the HHO generator is >= the power required to lift an extra column of gas at the same rate it's being generated.

I can't show that because I don't know the relationship between electric input power and pressure in an HHO generator. And you can't deny it because you don't know either. However by assuming conservation of energy we can conclude that the requirement in my previous paragraph holds.




Quote:

why would I go to all the trouble to just have a incompetent engineer work all the efficientcies out of the system , no I wouldnt hire any engineers.

sorry !
.


Then what? Go to the grave taking the greatest discovery in human history? Something I noticed about perpetual motion is that nearly every proponent of it creates his own idea. I've never seen somebody take another person's concept and promote it, try to prove it or build it. You came up with this bouyancy/water wheel idea yourself instead of just repeating what others have said. So equally, nobody else will take your idea and develop it. Unless you finish it yourself it will be lost and ignored forever.

Are you happy for it to be lost?
Will you commit your life to this work?
Or do you accept that it may fail?

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paul Offline OP
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Kallog

Quote:
we still have to push it into the pipe against the back-pressure of that column of air, which takes the same power as lifting the column of air itself, and in fact is what we are doing.


LOL

if the container at the bottom that is about to have the water in
it converted into HHO was at the top previously then it would have water and a small amount of air in it already and no HHO.

but its much simpler to use the pipe rather than the wheel , why you might ask?

because the pipe would cause less frictional loss than the wheel.

so I guess that removes your next straw.

using 2 pipes
(P1) to hold water, as it falls to the bottom.
and
(P2) to hold HHO as it rises to the top.

the water flows through a hydraulic motor at the bottom of (P1).
then enters the pipe (P2)that the HHO is in.

the water flows into the pipe (P2) via gravity.
water flows into pipe (P2) an into the HHO generator
located inside (P2)

pipe (P2) does not have an end on it at the bottom.

and the HHO is generated just above the bottom of the pipe
we have all seen that the HHO in all the videos floats upwards
in atmospheric pressure (1 atm) and there is no hardship placed
on the process due to atmospheric pressures or the buoyancy of HHO in air.

Im not sure that there wouldnt be some type of hardship placed
on it due to planetary alignment , or the coriolis effect

but I do know that HHO will float upwards in 1 atm pressure and that cannot be denied.

and I do know that water will flow down hill , that also cannot be denied.

now at the top of pipe (P2) the pipe is feeding HHO to
the power plant.

and the HHO at the top of (P2) will be slightly pressurized due to the buoyancy or density of HHO in air so it will
flow through to the power plant without undue strain or hardship.

dont believe me , try putting a glass held upside down over the top of a HHO generator , the HHO gasses will rise and push the air out of its way as it fills the upside down glass with HHO.

the power plant at the top will deposit the water from the power plant into the top of pipe (P1) which is open at the top to the atmospheric pressure.

now you could say that the water in pipe (P1) will evaporate
and that is true , some water will evaporate.

but if I add the amount of evaporated water at the bottom
of pipe (P2) then I dont need to lift it to the top , so no big deal as the company or home probably already has a 40 psi water
supply.

but you should also keep in mind that it rains these days
and a rain capture tank could be located at the top also !!

you should know by now that I dont believe in the energy conservation belief system that you believe in , as I know better.

I prefer to use the common sence method vs the theoretical
methods when determining if somethng will or will not work.

and just because you cant buy a FREE ENERGY system does not mean
that they cannot be built.

.



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Originally Posted By: paul

and the HHO is generated just above the bottom of the pipe
we have all seen that the HHO in all the videos floats upwards
in atmospheric pressure (1 atm) and there is no hardship


I think you ignored everything I just wrote. Generating HHO at 1atm uses more energy than generating it at lower pressure. You've said that yourself, and it sounds very reasonable. You find some evidence that it's wrong and you might have something.

I know you understand the idea that creating a bubble of gas lifts up the column of air above it. That consumes all the energy you might recover floating that bubble all the way to the top of the atmosphere. In other words dropping the column of air back to where it was.

Sure you can make the pipe higher and recover more of the wasted energy. At best you can approach recovering all of the energy by building a tower as high as the atmosphere.



Quote:

big deal as the company or home probably already has a 40 psi water supply.

but you should also keep in mind that it rains these days
and a rain capture tank could be located at the top also !!

If it depends on those things then it's not free energy, so who cares? Anyone can make a home hydro generator anyway. If it doesn't depend on those things then don't muddy the waters by mentioning them. They're a red herring.


Quote:

you should know by now that I dont believe in the energy conservation belief system that you believe in , as I know better.

I prefer to use the common sence method vs the theoretical
methods when determining if somethng will or will not work.

and just because you cant buy a FREE ENERGY system does not mean
that they cannot be built.


Show me any evidence of violation of conservation of energy. You can't, so why believe in it?

Show me any free energy machine that's ever been built. You can't, because none exists to public knowledge. So why believe in it? Why not believe in magic carpets or little green men on the moon?

Common sense is the antithesis of science. By definition common sense can never lead to any new discoveries. You're always constrained by limited personal experience. Theoretical science can go much further than the feeble human imagination. That's why the Earth now goes round the Sun, instead of the common sense alternative that people once blindly believed.

But again, why aren't you building it? You still don't appreciate the spectacular value it will have? I can understand not building a rainwater electric generator - it's pretty mundane and only marginally useful. But free energy will SOLVE ALL THE WORLD'S BIGGEST PROBLEMS FOREVER!!

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Kallog

the below system and its energy input and energy output (recovery) is opperating
durring a time frame of 1 minute.

if Im burning 1800 lpm at the top of the system.
and generating 1800 lpm at the bottom of the system
then I am putting 1 lpm of water in the pipe at the top each minute.


Michael Faraday has shown that 1 litre of HHO gas is produced
in one hour using 2.34 watts for a period of 1 hour.

2.34 watts X 1 hour = 2.34 watt hours each litre of HHO.

1800 x 2.34 watt hours = 4212 watt hours

if I use the 4212 watt hours in one minute this is my energy input



---------------------- ENERGY IN PER MINUTE --------------------------
252720 watts / 252720 joules / 4212 watt hours

---------------------- POWER IN PER MINUTE -------------------------

252720 watts / 252.7 kw
-------------------------------------------------------------------------


that part is now out of the way !!!!! we know our input energy.

---------- NOW FOR THE GRAVITY PART ------------------------------------


1 litre of water weighs 2.205 lbs.

the volume of 1 litre of water = 61.02374 cu/inches

if Im using a water pipe that has a inside cross sectional area of 61 sq/inches
then the water level in the water pipe will only drop 1 inch every minute because
I am only allowing 1 litre of water to pass through the hydraulic piston at the
bottom each minute , however I am putting back 1 litre of water in the top each minute.

to keep this simple as I like to do , we know we only have 1 litre a minute to work with

and I have to recover 4212 watt hours each minute to balance the recovery system out
(energy in = energy recovered) LOL.

so I will use a piston at the bottom and stroke it out once each minute.

if the piston stroke is 12 inches then the piston surface area will need to be
61.02374 / 12 = 5.002
5.002 sq/inches


this way every minute when it strokes it will remove 1 litre of water
from the water pipe at the bottom.

and when the piston strokes it will provide a force that can be applied over a 12 inch distance.

if the water pipe is 1000 ft tall , the water pressure at the bottom of the water pipe
will be 1000 X .433 psi = 433 psi

433 psi X 5.002 piston area = 2165 lbf applied to the piston
and applied for a distance of 12 inches or 1 ft in 1 minute = 2165 ft-lbf/minute

1 watt = 44.253 ft-lbf/minute.
2165 / 44.253 = 48.99

so if I apply the 2165 lbf over the 12 inches distance in one minute
I would recover a constant 48.99 watts during that one minute.

since my energy recovery system recovers 48.99 watts per 1000 ft in height
the ratio of energy recovery per foot in height would be
48.99 watts / 1000 ft = 0.0489 watts per ft in height.

I need to recover 252700 watts each minute as that is what it cost to produce the HHO.

so

252700 / 0.0489 = 5167689.16 ft high.

or 978.72 miles

since the energy out (recovery) is 252 kW/minute the energy that can be produced at the generator would be apx the same.

but the above is a way to produce more energy than you need , it would be much shorter if you were
focussing on just getting back the 22% you were loosing in normal water - HHO conversion.
and if its only 10% like you say , then thats even better.

and of course not everybody needs a 252 kW/minute power system.

the ISS is apx 199 miles high.

so that puts it right in the 22% range.

if I build a system this tall ((( ISS ))) ((( MARS ))) hint hint nanocarbons maybe !!!

I have recovered all of the energy I used to generate the 1800 lpm HHO

plus I still have all the energy that I produced at the top using the 1800 lpm HHO as a fuel in the power plant.

and if there are any inefficientcies found , just add a foot or so in height to compensate.


Im sure all of the stuff you write is important to you , and I do read it all , and you have found many
interesting obstructions that need to be considered , I just only reply to the stuff that you write that
applies to the discussion , I try not to focus on the energy conservation beliefs you have that you have
been taught and the what ifs that wont really accomplish anything.



the important thing about this is that it shows that the current usage of thermodynamics is a load of CR@P.


.


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Excellent calculation! That really does cut through all the talk and get the point.

However! :P I spot a serious fault. HHO is only bouyant in air, not in vacuum. So it can't float up past the atmosphere.

Assume optimistically that the top of the atmosphere is at the ISS (200 miles, there are some air atoms at the ISS). Your machine can recover 51kW compared to 252kW used generating the HHO. That's 20% of the energy recovered.

Hmm. That's a curious coincidence. Do you suppose the commercial HHO plants have about 20% loss because they're operating under a column of air that could recover 20% using bouyancy? That suggests this machine is pushing us right close to 100% HHO generation efficiency.

To show that it goes over 100% I think you need to find an existing HHO generator with more than 80% efficiency at 1atm. And also find out how high HHO will really float. I imagine it can't reach the ISS without somehow losing its usefulness.


---
Quote:

the important thing about this is that it shows that the current usage of thermodynamics is a load of CR@P.

So you've said several times, but still haven't shown it.

I understand that you don't trust conservation of energy. So you'll notice that I only referred to it tentatively for things I wasn't sure of. None of my reasoning depends on it, because doing so would be circular. If I depended on it I would simply say "E_out > E_in, violates 1st law, goodbye".

Well of course we all do assume the 1st law holds sometimes. But those cases are a bit too trivial and obvious to mention.



By the way, mixing imperial with metric, and miles with feet is the opposite of keeping things simple!!! Hehe that's why I don't check your calculations thoroughly. Too complicated :P

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Additional data:

Below the turbopause, at 62 miles, the different gasses of the atmosphere are well mixed by turbulence, and not stratified by density.

If we assume that the relative density of HHO and air doesn't change with pressure, then we should expect that HHO will be bouyant all the way up to the turbopause, and somewhat beyond.

So I think 62 miles represents a _lower_ limit on the maximum height of the tower. At that height we can recover 16kW from the 1l/minute of water. That's only 6% of the power used to generate the HHO so this result isn't conclusive either way.

I suppose what we really need is to find the altitude at which the density of the atmosphere is equal to that of HHO at the same pressure. That would be the upper limit on tower height. Beyond that the HHO won't be bouyant anymore.


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Kallog

what I said about the use of thermodynamics is true.

is physics becomming more like a religion these days?

for some reason I really think you know that the following
is true , yet you didnt seem to say anything about it.

if you reduce the scale of everything by say 100
then the 978 mile height becomes 9.78 miles
the 252 kW becomes 2.5 kW and so on.

you could even reduce the scale even further.


it would be entirely possible to make a floating platform at a 9.78 mile height , and using nanocarbon pipes that are supposed
to be stronger than steel , should make this entirely feasible.

now I know that the weather would present a problem but just the fact that this would work voids the first law as you use it.

but as I use it it will never be voided.

I could build something that could sit in your yard and it would be the size of a small swimming pool and it would produce HHO 24/7 and would power itself , and your home , and charge your electric cars batteries.

or you could just put one in you car.

you could put one in a airplane.

it would work in outer space , or on the moon , or on
mars... hint


but getting funding for anything that produces or saves any
large amount of energy is impossible because the people who supply the funding rely on people who have been taught wrongly for their opinions on the invention or process and they simply use the first law as their knowledge base and that pretty much dissallows any energy savings or free energy production.

I have tried many times , and am always shot down by thermodynamics.


even though thermodynamics is not used correctly its wrongfull usage is the correct way to ensure destruction.

.


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Originally Posted By: paul
Kallog
what I said about the use of thermodynamics is true.

is physics becomming more like a religion these days?

Whatever other people might be doing isn't relevent. They're not holding you back, they're working on their own things. Your intention is to disprove part of the theory. So you don't have to depend on that theory. That's fine. Scientists are doing that all the time with lots of theories. Go for it. I'm also not depending on thermodynamics when I talk to you. Except of course the obvious cases that we both assume are true. For example, the density of a liter of water doesn't spontaneously double if you leave it sitting under the full moon and sing a special chant.


Quote:

if you reduce the scale of everything by say 100
then the 978 mile height becomes 9.78 miles
the 252 kW becomes 2.5 kW and so on.

Correct. And the 1 liter of water becomes 0.01 liters.
then the water flow rate down the pipe reduces by 100.
And the reduced height causes the pressure at the piston to reduce by 100.
So the power generated by the piston reduces by 10,000 to 25.2W!!

You're welcome to repeat your calculation for any other set of numbers if you think it'll improve the result. You can even use an imaginary planet with a different atmosphere, different gravity, etc.


Quote:

I could build something that could sit in your yard and it would be the size of a small swimming pool and it would produce HHO 24/7 and would power itself , and your home , and charge your electric cars batteries.

Go for it. But it's already a big step if you just show theoretically that it would work. That costs nothing except your time. But to persuade someone you have to counter every single critisism that anybody makes. It would also help if you explain how everybody else managed to be so wrong for so long. People will want to know that. I mean specifically how - what mistake was made in what experiment by who? Where did somebody forget to divide by 2? Where was a result extrapolated to an untested area? that sort of thing. Point your finger on the actual mistake so everybody can clearly see it. Maybe gravitational potential energy is excluded from the 1st law? Identify the actual problem. Then you can easily test your claim.

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paul Offline OP
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Kallog

yea I did forget to reduce the lenght of the pistons stroke by 100 , and I havent calculated the reduction in energy input
either so I might do that later but thats really not necessary.

so rather than putting up another example to use I think it
might be best to use your idea.

Quote:
You're welcome to repeat your calculation for any other set of numbers if you think it'll improve the result. You can even use an imaginary planet with a different atmosphere, different gravity, etc.


as thermodynamics should apply anywhere correct?

and since the 978.72 mile high example would work
if it were sitting in a hole here on earth.

we can still use the earth.
and the same atmosphere.
and the same gravity.

remember this is only an example just to prove my point
that the current use of thermodyamics is BULL$#!T.

not that its meant to actually build anything.

so if the 1 lpm 252 kW example was sitting in a hole here on earth.
and the hole was perfectly insulated from the heat of the magma.
and it had a roof over it to keep rain out.
and the great purple rinocerous was no where in the vicinity.

it would work if built in a hole on earth , correct?
correct!!!

your only reasons given previously to its inability to function was the fact that there is no buoyancy that high up.
does buoyancy work that far down?

if we venture out beyond the earth would we need to adhere to
the current usage of thermodynamics as it is supposed to apply anyplace , or should we throw it in the trash can where it belongs.


.


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Yea of course thermodynamics works everywhere - as far as we know.

The hole in the ground suffers from a new problem. Your calculation depends on Faraday's relationship between electrical energy and volume of gas generated. That was done at 1atm pressure. At the bottom of the hole there's higher pressure, and you've said yourself that'll require more energy. Maybe it's not so much more that it breaks the idea, but unless you find out, you can't assume it's OK.

I know it sounds like I'm just making up a new silly excuse every time. But that's the nature of perpetual motion ideas - robbing from Peter to pay Paul, then robbing from John to pay Peter, then ... until you start to wonder if there's some fundamental roadblock that causes nature to always preempt your intentions and defeat them.

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paul Offline OP
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WOW

I guess it just wont work unless you can make it shorter
to remove the pressure differences at depths.

let me think about this today to see what I can come up with
to remove the pressue obstructions.

there is another system I would like to discuss with you
later that involves an added process in apx the same situation.



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paul Offline OP
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Kallog

I found a way to do this today but it does produce the HHO
under pressure only its at a 30 ft depth not 978 mile depth.

LOL

I either need to find the energy needed to produce HHO
under pressures or I need to find how much energy is required
to create a vacume.

but I will do that later , it shouldnt be hard to do as I
still have 57 kW left over from 309 kW recovered in the minute.

I still think this is a workable example because after the minute
I have 57 kW and 1800 litres of HHO have been used in the power plant.











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Hmm using some of that 20% recovery rate is going to be tight. But I look forward to it.

Energy of a vacuum is easy. If you open up a 1m3 cube of vacuum on the ground, you're effectively lifting the column of air above it up 1m. So the force required is 1m^2 * 1atm pressure. Energy = force * distance:

E = 1m^2 * 100,000 N/m^2 * 1m
= 100,000 N m
= 100,000 J
= 27.8 W-hrs
= 1668 W-minutes to generate 1m^3 of vacuum or
1.67 W-minutes to generate 1 liter of vacuum or
3000 W-minutes to generate 1800 liters of vacuum.

A thought I had was to use an atmoshpere made from a gas that's just slightly denser than HHO. That way the atmosphere can be thicker while having the same pressure at sea level. The HHO would still float all the way up, especially if there's only one gas in the atmosphere so it doesn't stratify into lighter gasses near the top.

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Kallog

Quote:
Hmm using some of that 20% recovery rate is going to be tight. But I look forward to it.


well the 1800 litres HHO cost 252 kW / minute if I remember correctly , then I recovered 309 kW / minute.

50.4 kW would be 20% of 252kW.

100% would just be making up for the lost 20% normally seen in HHO production or if I only recovered 50.4 kW.

so I have 57 kW / minute to use for the vacume or the extra watts needeed to convert HHO in the higher pressure water before I go below 200% efficiency.

so for now I will wait until I find out how much more
energy will be required to produce HHO in pressures or
in a vacume.

but I already think that the vacume wont require much more
at all.

so right now Im at 200% efficiency plus 57 kW per minute.

btw.

thanks for the vacume info , how would you calculate
a 5 inch vacume?

vacume is measured on a guage by inches , at least thats
what my guage reads , I suppose I can just guess the force I apply to the vacume pump and measure its piston area x stroke , then measure the volume of air inside the flask I am throwing the vacume on since the guage will tell me what the vacume is , I should be able to make a close estimate this way.

it seems that the water wants to boil at apx 22 inches of vacume so certainly I wouldnt need that much and it takes several pumps with the hand pump to achieve 22 inches of vacume.

I say boil but its not boil like water and heat , its like
hydrogen comming out of the water using just a vacume.

anyway I think that the extra 57 kW / minute sholud be capable of doing this wouldnt you think?











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Inches could be inches of mercury. 1 atmosphere is about 30 inches of mercury. But it would go backwards for vacuum so atmospheric pressure is 0in. and complete vacuum is 30in. What I worked out was for a complete vacuum, even tho we can't quite get there, and the pumps are probably horribly inefficient, but that's just a technological problem.

That would mean 5" takes 5/30 = 1/6 of the energy needed to get complete vacuum on the same volume chamber.

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