Gravity Mill - any comments to this idea?

[2tiger]

Hello prajna

Water at sealevel has a force of 14.5psi or 1.02kg acting on it.  If I have a 'U' tube such as the one in your drawing that has an internal diameter of 3.57cm it will hold 1 litre/m of tube.  Let's assume that there is enough water in the tube so that the bend at the bottom is under water.  If I then pour 1 litre of water into the tube then the water level in the tube will rise 50cm.  Equally, if I put a 1kg weight in one leg the water level will raise 50cm (since 1 litre of water weighs 1kg).  Is that what you were trying to say?

Yes this is right. But only if the U tube has two legs of the same diameter. As the ratio between the legs is 1:2 you need to pour 2 liter, if you want the water to rise 50 cm.

What will happen if there is just 40cm of height of tube over the level of the water and I put a weight of 1kg in one leg? 

Yes - this is what I wrote. Remember (look at the picture 1 Marble 1cm in diameter):

As you can see the 50 % more weight on the right side is only able to lift the blue marble on the left side half his height - 5 mm. If the marbles were water, the water will rise 5 mm too. But if you want the water to overcome the 5 mm so that it can flow out, you have to choose a tube that is 4 mm (over sealevel) high OR you have to make the shuttle 50%+10% = 60% heavier (by same volume of water)!

The water will flow out of the other leg (because we know that 1kg will push the water up 50cm) and the weight will continue to sink, pushing even more water out of the tube until it gets to the bend at the bottom.  It will have pushed nearly half of the water out of the tube (considerably more than half a kg of water) even though the weight I put in is only 1kg.  This is how ELSA works when the weight is going down.

Yes - this is the conclusion of that, but only in a 1:1 ration between the both legs of the U tube!
And therefor you will need 2kg to overcome 100 cm, 3kg to overcome 150cm and each time subtracting 10 cm of the tube over waterlevel, so that the water can flow out!!
In the case of a 1:2 ratio, you will need 4 kg to overcome 100 cm, 6 kg to overcome 150cm.
In the case of a 1:5 ratio, you will need 10 kg to overcome 100 cm, 15 kg to overcome 150cm.
...and so on...

And stefan here is the answer to your question, how high will the water rise over sealevel.
I do the math for you.
Imagine a U-tube 100 cm under sealevel, one leg of U-tube over sealevel and diameter ration between the legs 1cm:1cm.
Now if the shuttle is 50% heavier than the water by same volume, the water will rise 0,4 cm.
In a 1:5 U tube just 0,008 cm.

Cu
2Tiger

[keytronic88]

Hi All, Maybe this should be under another topic: but here is another water gravity wheel.
http://www.raska.info/tocak2/Invention.pdf#search+%22vukosavljevic%20perpetuum%22

Regards,
Keytronic

[tbird]

hi prajna,

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the only thing you would have to do is divide by 2 at the end.
Surely just subtract 1 at the end (or am I tired and confused this morning too?)  Whatever the absolute pressure is in the container the gauge pressure will read 1 atmosphere less because the gauge is zeroed to 1 atmosphere, yes?  

i have to say, i'm sorry.  i didn't verify my statement "works every time".  this is another example of how easy it is to get confussed with absolute.  to check your formula, i changed the pumping tube depth to 20m.  the answer is right.  it comes in at 6 atmospheres.  in order to get gauge pressure, if you subract 1, it would be wrong.  it would be 2 times the 1 because you had to double it as well as the water pressure.  so instead of dividing by 2, subtract 2.  not being a program writer myself, i can't tell you the best way to write this formula.  but to get to the weight needed, all you need to know is the pressure and area the force needs to be applied to. assuming we only want to double volume from static state at depth.

Ok guys, I think that is as far as I will go with this design of ELSACALC (unless tbird wants me to add metric to imperial conversion in this version).  Next I will work on John's double acting ELSA so that we can calculate the output from that.

it would be a wonderful addition for me (probably Mr. Herring too).  it's your program, you have to do the work (i no it takes a bunch), so it's up to you.

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why not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly.
Because I want to allow the user to experiment with the effect of changing the height of the head.  All of the inputs are there so that the user can experiment with the effect that changing their values has on the various results.  It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' and the head height is the main determinant for that.

The user doesn't have to provide a value for this, since it is a default.  I understand what you are saying though: it would be nice to have the most efficient system specified as the defaults but calculating the most efficient system is not a trivial task.  Perhaps later I will change some of the default values.

can't you do both? from just above " It seemed logical to me that one of the first things you might want to decide is 'how high can I pump the water?' ".  imagine an head pipe the size (diameter) of a toothpick with 50kg of pressure behind it.  how high is that, max?  no, because you can always divide that into a smaller size.  not practical.  thus my suggestion of a percentage.  i didn't mean to limit what the user could use, just start there and still be able to change the value.  basicly the same as you are doing, but with a footnote like "this value is x% of pumping pipe diameter".  it would be your default.

I would very much like to come up with some elegant way to recompress the shuttle for ELSA though.  All of the rest of John's design is nice and simple but the recompression methods are all a bit ugly and Heath-Robinson.  Perhaps that is why he has had difficulty introducing the concept to the world.

i agree.  just be careful not to get too cute.  like you said something about an electric control for....  having lived on a boat for as long as i have, you learn to keep your electric devices as far from water as possible.  you only mix them if there is absolutely no other way (if you don't want to be doing without that device from time to time until you can repair the water damage).  john had an idea (shows in drawing #47) that might be option for those electric things you talked about.  simple magnets.  don't think the drawing covers all you wanted to do, but you're a clever guy, keep going down that path.

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raising the delivery height increases the energy stored.  wouldn't time have to be considered somewhere?
Certainly.  At a later date I would like to introduce fluid dynamics to the mix and friction so that we can calculate the stroke speed.  That will allow proper output values to be calculated and we will know if you really can power your house with one of these in your yard.

when you get there, my post will probably slow down.  they may only consist of questions.

keep up the good work!

tbird

[tbird]

2tiger,

you must be overlooking (at least) the fact we don't use the same size diameter pipe above water.  true it's slower delivery, but it all will go there except what is left in the head pipe.  since our examples (so far) are based on 10m depth, there's plenty of water lifted to do work.

tbird

[2tiger]

Hello Tbird
It doesn?t matter if it is 10 m or 100 m depth, because the watercolums (without shuttle) are canceling its weight eatch other out thrugh the hydrosttic pressure. The only thing that takes effect (unbalance) here is the shuttle with 50% more weight (comparing with same volume of water). As result you?re still lifting half of the shuttle height.
Is the shuttle 1m high then you can lift 50 cm over seawater, but the tube has to have a little bit less (49cm) so the water can flow out.

If you choose a lower diameter for the tube over sealevel, let us say the half diameter of the tube under water, then you will have to double the pressure by making the shuttle two times heavier.
ONLY in that case you were able to double the height over sealevel.

That is the hydrostatic paradoxon!

By
2Tiger