Author Topic: Gravity Mill - any comments to this idea? (Read 101871 times)

prajna · « **Reply #240 on:** August 28, 2006, 03:13:20 PM »

tbird,
my approach is to get a rough idea of how something works and then work all of the details out for myself. That means to ignore the manual completely and, though that might sound dumb, it means that I end up with a deep understanding of how something works - much deeper than if I had simply read about it.

I've updated the calculator a little. Nothing significant changed but I have added a picture to make things a little clearer.
http://declarepeace.org.uk/energy/elsa.htm

Quote

maybe you just left out another step that would leave you with enough compress air.

I think I straightened this out in a recent posting:

Quote

At the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it.

Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?

tbird · « **Reply #241 on:** August 28, 2006, 03:46:35 PM »

Quote from: prajna on August 28, 2006, 03:13:20 PM

tbird,
my approach is to get a rough idea of how something works and then work all of the details out for myself. That means to ignore the manual completely and, though that might sound dumb, it means that I end up with a deep understanding of how something works - much deeper than if I had simply read about it.

I've updated the calculator a little. Nothing significant changed but I have added a picture to make things a little clearer.
http://declarepeace.org.uk/energy/elsa.htm

Quote
maybe you just left out another step that would leave you with enough compress air.

I think I straightened this out in a recent posting:
Quote
At the bottom of the cycle air is released from the bottom of the shuttle (think of it as an upside-down hollow piston) into the pumping cylinder until the pressure is equal inside the shuttle and below it.

Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?

had a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.

so there is a small passage outside the shuttle (still inside the tube) to valve 3?

your shuttle does get pretty close to elsa design. if i see your shuttle right, the bottom of the tube will be open? to allow the shuttle to descend without moving water up threw exit pipe?

tbird

ooandioo · « **Reply #242 on:** August 28, 2006, 03:53:43 PM »

Quote from: prajna on August 28, 2006, 03:13:20 PM

Aah, I see your confusion. The displacement area is below the shuttle (not the reservoir inside the shuttle). I let the air out of the displacement area but don't let out the air I have just pressurised into the reservoir. Is that clearer?

Ok, now its much clearer and actually I think its the best design we currently have.

prajna · « **Reply #243 on:** August 28, 2006, 04:21:24 PM »

Guys, for the calculator the shuttle design is even simpler than the previous one. Take a look at the picture on my calculator site. It is just a hollow piston into which we can compress air. At the bottom of the stroke the air is released from the piston (via a valve, presumably) underneath the piston. This means that the compressed air that was inside the piston is now partly inside the piston and partly underneath it. But now it has expanded from the volume of the piston (actually slightly less because we have to take into account the thickness of the piston's walls so its internal capacity is slightly less than its displacement) to the volume it will occupy at the depth we release it (10m in my default example).

Does this all make sense?

prajna · « **Reply #244 on:** August 28, 2006, 04:29:10 PM »

Quote

had a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.

It is heavy, tbird. That is because you have a much taller shuttle than you need. What kind of head are you lifting? If it were 1kg (2.2 pounds, I think) then you would need a much smaller shuttle. Tell me the head sizes as well as the pump sizes and I will tell you a reasonable shuttle height.

How much does one cubic foot of water weigh?

ooandioo · « **Reply #245 on:** August 28, 2006, 04:32:57 PM »

Allright, the idea is very good - if we are able to seal the shuttle sides so that no air is able to escape.

prajna · « **Reply #246 on:** August 28, 2006, 04:57:21 PM »

We can do that, as per my previous design, once all the figures have been calculated but it is simpler to consider it as a simple piston for now.

hartiberlin · « **Reply #247 on:** August 28, 2006, 05:55:45 PM »

Hi Pranja, good work with your calculator.
Well, one thing:

So we can pump 77754.42 litres on each stroke
using Ãâ‚¬ r2 * (height of the pumping tube - height of the shuttle)

So now your are subtracting the height of the shuttle ?

Is that with the displacement area full of air or without it ?

Maybe you make better the main tube x meter longer when you
ask for the deepth, so the tube will be automatically 20 cm longer than
10 Meter, if one chooses 10 Meter deepth...

Then it will not be too confusing...

2. Also it would be better, if you draw your picture this way, that
you attach some open buttom case for the air displacement area,
otherwise one can not easily see, that air is expanded from the shuttle
down below the shuttle into the displacement area.

Many thanks for your hard work.

prajna · « **Reply #248 on:** August 28, 2006, 06:07:17 PM »

Stefan. I have changed it again. Now it calculates the pumped volume by subtracting the volume of the total air required to lift the shuttle and the head from the volume of the pumping tube. To save questions, this is the volume that the air will occupy at the bottom of the pumping cycle including the displacement of the shuttle.

The height of that air in the pumping cylinder will become important when we need to figure out how close to the bottom of the pumping tube the shuttle can go so I will also include the height figure in the results. It also tells us the height of the shuttle including a skirt to contain the expanded air (which you suggested I show on the drawing).

tbird · « **Reply #249 on:** August 28, 2006, 06:30:22 PM »

Quote from: prajna on August 28, 2006, 04:29:10 PM

Quote
had a short look again at your program. with 15cm (+or-6inches) diameter and 30cm (+or-12inches) high (sounds like less than 1 cubic foot (64.7lbs)) the result is 53.01 kg. doesn't that sound heavy to you? maybe you need to use internal radius instead of internal diameter.

It is heavy, tbird. That is because you have a much taller shuttle than you need. What kind of head are you lifting? If it were 1kg (2.2 pounds, I think) then you would need a much smaller shuttle. Tell me the head sizes as well as the pump sizes and I will tell you a reasonable shuttle height.

How much does one cubic foot of water weigh?

unless i really don't know my metric values, 15cm diameter by 30cm high is only about (not even, 6x6square has more area than a 6"diameter circle) 25% as big as 1 cubic foot. so 25% of 64.7 pounds (16.175 pounds or 7.35kg) doesn't sound like 53.01 kg (116.6 pounds). what am i missing?

Quote

so there is a small passage outside the shuttle (still inside the tube) to valve 3?

your shuttle does get pretty close to elsa design. if i see your shuttle right, the bottom of the tube will be open? to allow the shuttle to descend without moving water up threw exit pipe?

didn't see an answer to this part of my post. should i quack?

prajna · « **Reply #250 on:** August 28, 2006, 06:46:07 PM »

tbird,

a cylinder of diameter 15cm and a height of 30cm has a volume of 53.01 litres

53.01 litres = 1.87203048 cubic feet so that is where your error is.

Unless I have my decimal point in the wrong place. I'll go check.

I have added a detail picture of the shuttle so that might answer your second question. Don't quack

prajna · « **Reply #251 on:** August 28, 2006, 07:21:54 PM »

Yes, decimal point in the wrong place. Sorry tbird the volume should be 5.3 litres = 0.187167734 cubic feet. I'll fix my calculator. Thanks for pointing it out.

I'd better check my pumping capacity too.

prajna · « **Reply #252 on:** August 28, 2006, 08:30:55 PM »

phew! I think I have decimal points in all the right places now. Can someone please check by calculating the volumes for themselves and comparing the results to my calculator.

Oh, and I have fixed it so that the javascript runs in DOM compliant browsers like Firefox too.

Time to add some compression calculations I think...

http://declarepeace.org.uk/energy/elsa.htm to save you having to find the link.

prajna · « **Reply #253 on:** August 28, 2006, 11:23:45 PM »

Announce, Announce...

ELSACALC now calculates the recompression requirements and excess water. Read it 'n weep ('cause we never worked on this earlier).

Anyone care to check my calculations?

http://declarepeace.org.uk/energy/elsa.htm

hartiberlin · « **Reply #254 on:** August 28, 2006, 11:33:32 PM »

Hi guys,
I think the ELSA system is the first system which can be calculated mathematically
to run and to have overunity.

Now I thought a lot in the last days, how to best build such a unit, but I must say, it is
very complicated with all the mechanics involved.
So I am thinking now about how to do it much more easily.

One way could be to just rotate the main water case by 180 degrees,
when the water has been pumped up and over the center of gravity
of the whole unit, so it can tilt by 180 degrees and the cycle can begin again.
Then you also don?t need to compress the shuttle anymore.

This is also not so easy cause you have to see how to get more water weight
over the center of gravity of the whole unit..

A second new idea is to just use a slow rotating water cylinder with
a fixed volume shuttle in it and every time the water cylinder has turned
by 180 degrees the shuttle with buoyant force is again at the ground of the cylinder
and will rise up again.

Think about a Plastic coke bottle which is filled with water to the top.
Then put a ping-pong ball in it and screw the lid tight.
Then rotate the coke bottle and you see, that the ping-pong ball
rises every 180 degrees from the buttom to the top
inside the coke bottle, if you continue to rotate the bottle.
Now if you have the a bigger and longer watercylinder and
have a bigger shuttle than a ping-pong ball you have real good
bouyant forces, which can pull via a thread on a generator axis
in the watercylinder, so you could generate quite some power
this way and couple it to a flywheel that keeps
the watercylinder rotating !
I guess this also could work. I will
calculate the energy you would win per 180 degrees lift.

Regards, Stefan.

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