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Author Topic: Gravity Mill - any comments to this idea?  (Read 100911 times)

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #150 on: August 26, 2006, 08:35:02 PM »
At seawater level the shuttles air volume has  now doubled, cause:
P1 x V1 = P2 x V2

so it is now 3200 Liters !
 Now we must release almost 2415 Liters from it, so it has got still less than 785,4 liters, so the 785,4 kg weight attached to it can pull it down completely...  Agreed so far ?
« Last Edit: August 26, 2006, 08:57:04 PM by hartiberlin »

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #151 on: August 26, 2006, 08:42:41 PM »
Now we move down to 20 meters and the airvolume will compress even further, cause we are now at 3 bar air pressure inside the shuttle ballon, due to the outside water pressure and the volume of the balloon has reduced to around 261 liters only... Agreed so far ?

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #152 on: August 26, 2006, 08:43:37 PM »
hi all,

just called Mr. Herring.  got his answering machine with a message from his wife.  i left message telling about this forum and my phone #.  hope he shows up.

tbird

bastonia

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Re: Gravity Mill - any comments to this idea?
« Reply #153 on: August 26, 2006, 08:48:35 PM »
fyi - Just sent John an e-mail ... also inviting him to chat ...

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #154 on: August 26, 2006, 09:09:26 PM »
Okay, now we need to refill the shuttle balloon isobarly ( at the same pressure)
at 3bar with about additional 1310 Liters again so it has a total of a bit more than 1570,8 Liters,
so to be able to rise it against its own 785,4 Kg attached weight and also against the 1 Meter water colum
it has to push beyond seawater level, which is another 785,4 Kg weight.

You see this is a lot of energy we must apply for the pumping !
To pump these 1310 Liters at 3 bar we need alone 109 Watthours of energy to do this !

So, this way it does not work.
We really must use a smaller pipe diameter just directly above seawaterlevel
so there must be no water column weight to be overcome and the water must
then be  sprinkled into an upper bassin via the pressure from the buoyant force,
just like a fountain.

Then we also can reduce the attached weight to the shuttle and have to pump less
air down at the 10 or 20 meter deepth.

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #155 on: August 26, 2006, 09:27:49 PM »
Okay, now this step by step analysation has helped a lot to understand the
thermodynamical principle behind it and all the PV diagrams of the air which  went through
these cycles.

Now to optimize the cycles we must do the following:

Have no watercolumn above seawaterlevel, so we reduce
the negative impact of the hydrostatic paradoxon.
Instead have a nozzle directly at the seawaterlevel
inside the tube?s top, so it works like a fountain
and sprinkles the water 1 Meter at least out into an upper
reservoir at 1 Meter height.
I wonder how big must then be the buoyant force on the shuttle
to get the water  pushed out 1 Meter high behind the nozzle ?
Are there any forumulas to calculate this ?

2. We should try to calculate the needed energies to do this with a shuttle,
that has really only 2 different volume states and does not expand
or shrink with water deepth pressure !

So I will try to do this now.

ResinRat2

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Re: Gravity Mill - any comments to this idea?
« Reply #156 on: August 26, 2006, 09:39:47 PM »
Hi everyone,

ooandioo and tbird, thanks for the info and kind words. I've been busy getting ready for my son's B-day party and am just getting back (snuck away from the wife) to reading the posts. A great deal of activity here today.

Hope tbird's and bastonia's attempts at contacting Mr. Herring work out.

Stefan, it doesn't seem like you read my first post, but check out this link to a drawing from Mr. Herring.

http://www.icestuff.com/energy/elsa/page_36_-_780.htm

It gives a simple way to recompress the shuttle. No electronics, no timers or switches. just a lever that can be weighted down by the water above.

If you get a chance read my earlier post.

Thanks everyone for all your efforts.

ResinRat2

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #157 on: August 26, 2006, 10:03:29 PM »
ResinRat2,
yes, let?s concentrate next on the recompression of the shuttle at the top.
To do this efficiently we really need to work this out with all the equatations.

I have to pause for a little, cause I have to take a bath and play with the tubes
and shuttle in it  and see, how the nozzle effect turns out and how much
force it needs onto the shuttle !

pese

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Re: Gravity Mill - any comments to this idea?
« Reply #158 on: August 26, 2006, 10:10:04 PM »
recompressing , also good to understand with:
http://www.icestuff.com/energy/elsa/page_29_-_770.jpg

GP

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #159 on: August 26, 2006, 10:22:03 PM »
Mr. Herring has drawn some nice graphics,
but the problem is, that you have a special air pressure at
each deepth stage inside the shuttle and a different water pressure around the shuttle,
that has to be overcome, when you want to compress or push out the shuttle volume !


This needs real energy to do so, as you also have to change the
air pressure inside the shuttle. To change the airpressure inside the shuttle
is also not less energy requireful than to do it
with just a pump.

And for the pump we already know the equatations.

If you want to use the lever torquearm-combination you also have to calculate in the
distance you have to move the other side of the lever torquearm to get the desired
pressure, so to calculate the required energy you need to calculate the
forces x distance. If you move the shuttle pistons just 10 cms to achieve the different volume,
you have to move the lever arm at the other side, maybe 50 cm to have a 5:1 force multiplication
and then the question is, if the water is high enough, so you can do this 50 cm movement ?

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #160 on: August 26, 2006, 11:21:07 PM »
hi andi,

sorry to take so long to get to this post.  been thinking about how to get stefan a better internet.   :D

you wrote:

"hartiberlin, thats what I'm saying. Fixed volumes will not work as the pressure is different at different sea levels. Maybe it should really be a static box with 2 volume states. Lets see, what your calculations are going to look like."

can't imagine what you mean by "fixed volumes".  also, what is a "static box"?  does it look any thing like the shuttle in this drawing?    http://www.icestuff.com/energy/elsa/

as far as having different pressures, that's ok.  did you realize the shuttle in the drawing is pre-pressurised at the surface, before it ever makes it's 1st run?  this pressure, when the shuttle is compressed so it will desend, will be 2 atmospheres (if we are taking it to 1 atmosphere, 10meters +or-).  30psi is a little more than that, but will make the shuttle expand to size, for sure.  to compress a shuttle that is based on a one cubic foot displacement, we'll say the top area is 12 inches square (144sq.in.total), would take 30psi times the 144 square inches which equals 4,320 pounds.  even if the shuttle is expanded with 15(+or-)psi at the surface, 4,320 pounds will compress it to the point where the air being compressed inside will be 30psi.  since we started with 15psi, the volume will be reduced by half.  back to the weight.  if you use a lever to hold the container that the compression water (weight) will be put in, for every foot out, less water will be needed.  if the lever was 5 feet long, the weight would only need to be 1/5th (864lbs) as much.  if you took it out 10 feet, 1/10th (432lbs.).  now at 10ft leverage we've reduced the amount of water needed to recompress from 7.67 cubic feet (4,320lbs.) to 0.767 cubic feet (432lbs.).  since we are goin down to 10 meters (+or-) and delivering water in both directions, we should have an excess of about 65 cubic feet (over 4200 lbs).  not bad for a 1 cubic foot unit, eah?  using the example from this drawing  http://www.icestuff.com/energy/elsa/, the height of the stored energy (water), (if you store it) could be as high as 72 inches using a 1 square inch exit pipe.

do you still want to only use the up stroke?

i have more about the shuttle, if you are not burnt out.

tbird

ooandioo

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Re: Gravity Mill - any comments to this idea?
« Reply #161 on: August 26, 2006, 11:51:36 PM »
This is really amazing. Some times ago I found elsa on the internet, but nobody would recognize my questions to it. Now its going to be the biggest thread in this forum...

tbird, yes, I'm coming back to the original elsa design. Using a shuttle box that can have 2 different volume states will be the right decision. But, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth. I think first we have to find out if the gained waterpower is able to compress the shuttle. Hartiberlin is on the right way.

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #162 on: August 27, 2006, 12:04:04 AM »
andi,

didn't you understand my last post?  do you think my numbers are wrong?

tbird

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #163 on: August 27, 2006, 12:15:33 AM »
andi,

btw,

Quote
But, as I said some times ago, the pressure is different at different shuttle depths and it doesn't make a difference if we compress the shuttle at or above waterlevel or in 20m depth.

with this shuttle, those pressure differences are not a factor.  when the shuttle expands at the bottom, it will be at it's max volume.  the displacement is the same all the way up.  this brings me to the "i have more about the shuttle" statement.  to take advantage of the pressure left over at the top of the cycle, we could build the shuttle to accomdate more expansion.  it won't deliver any more water in that cycle, but it will do it faster.  so for a given time, we will have more engery (water) available.  the recompression energy will stay the same too.

tbird

prajna

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Re: Gravity Mill - any comments to this idea?
« Reply #164 on: August 27, 2006, 02:09:43 AM »
Hi guys,

I'm hoping to simplify things a little with the attached drawing of a shuttle.  The displacement area can be full of water or air (depending on whether the shuttle is going up or going down.

Let's work with just one kilo of water and a shuttle weight of 0.5 kilos to keep things simple. We must also consider the volume of our shuttle because it will also displace that same volume of water, let's make things easy and assume it is 0.5 litres. The displacement volume, then, must be 1 litre  - 1 litre to displace 1 kilo of water plus 0.5 to displace the weight of our shuttle minus 0.5 because our shuttle already displaces that much water when the displacement area is full of water. The weight and volume of our shuttle cancel each other out if they are the same.  If the weight is less than the volume then it floats and if the weight is more than the volume it sinks.  Ok so far?

The displacement volume will be the same at any depth but, as Steve pointed out, it requires greater air pressure to displace 1 litre of water as we go deeper because we have to displace that 1 kilo of water plus the weight of a column of water above it (just as at sea level the air pressure is approx 14.7psi - or approx 1 atmosphere - because we are measuring the weight of the column of air above it, from sea level all the way up to the top of the atmosphere.)  At approximately 10m below waterlevel the weight of the column of water is the same as the weight of a column of air the height of our atmosphere!

What our shuttle will do is to contain air in the reservoir sufficient to displace 1.5 litres (i.e. 1.5 kilos, see why we love metric) at a depth of 10m That is 1 litre of water plus the 0.5 litre volume of our shuttle because we will be keeping that volume in our reservoir when we have filled the displacement area. At 10m we will require 3 litres of air in our 0.5 litre reservoir so it must be at 6 times atmospheric pressure - if my logic is correct. That should be 14.7 * 6 = 88.2psi = approx 6.2 kilos/cm2.

We can discount 1 litre of air when it comes to recompression of the reservoir because we still have that from the previous cycle. This leaves 2 litres to recompress = 2000cc. If we have a plunger with a cross section of 1 cm2 and a length of 20m then a weight of 6.2 kilos will compress the air in the cylinder to that pressure.  If we only have 0.5m to compress in then we will have a cross section size of 40cm2 and require a weight of 248 kilos.

The quantity of water we can pump depends on the area of the top of our shuttle.  If it is 20cm2 then 1 litre will be 50cm high in the pipe and we can therefore pump 20 litres 0.5m above the water level. Hmmm....

Is my maths wrong or did I miss something or do we have less energy than we thought?

The proposal for the valves, by the way is that V1 is used to fill the reservoir with compressed air, v2 lets the compressed air into the displacement area when the shuttle reaches the bottom and V3 lets the air out when the shuttle reaches the top.

Forgot diagram.  See next message.
« Last Edit: August 27, 2006, 03:38:52 AM by prajna »