Gravity Mill - any comments to this idea?

[2tiger]

Congratulation Stefan!
Now I am sure that the E.L.S.A. device won?t work as you assume at the beginning of this thread.
But your device will work if you add some things.    

If you let your device lift up with water and tube and try to hold the tube in this upper position (look at my drawing).
Then you have only to open the valve in order to unbalance the system. Some water will flow out and the lowered weight will let the swimmerboddy raise up till he reaches his upper position (floating).

Again CONGRATULATION accidently you have find the right setup.

The best thing in your setup is that it doesn?t have to be a large tube, because the lifting work begins right under the watersurface, the depth will equal to the hight of the lifted watercolum.

Good job!

Now you have to do some maths to calculate if it is still OU.

By
2Tiger

[ooandioo]

Hi 2tiger.
I think you misunderstood something about elsa. Your second drawing is allright, but that was the idea from the beginning. Imagine a large sea or pool and one long pipe going 10m under water and 1m above the water. The pipe is surely fixed in this position.
The transporting shuttle is able to freely rise up from 10m under water in the pipe and transport all of the water above it in an upper basin, 1m above sealevel if it has the righ boyancy.
We are now researching some math about how we are going to compress the shuttle, or the air inside the shuttle, so it can increase its volume 10m under sealevel.

Andi.

[prajna]

Glad you like it guys.  There is a lot to answer - mostly because the page is not complete yet.

@Stefan - yes I will be adding recompression calculations, COP and metric to imperial conversion.

@tbird - the shuttle weight is calculated as that weight required to sink the shuttle (i.e. balance the volume of the shuttle.)  The volume of the shuttle can be set by changing the pumping cylinder internal diameter or the shuttle height (the most logical thing to do.)  Yes, I have based everything on an ideal shuttle - that is a cylinder that has infinitely thin walls.  There are many improvements that can be made such as to choose material for the shuttle so that we can calculate the internal volume of the shuttle e.g. if the shuttle is made of lead the walls will be Xmm thick and the capacity will be Ycc less than the displacement volume.  We can add a friction calculation to tell us how much energy will be consumed by the seal between the pumping cylinder walls and the shuttle.  We can add viscosity calculations for the water in the header tube. etc.

the first thing you do is "head of water in the feed tube".  this weight is nice to know, but i think just as (if not more) important would be to know what the max diameter of that pipe could be for a given height.  and maybe if given a pipe diamter, what max height could be.

shortly after you get to the shuttle.  it's a bit unclear if you are referring to expanded or compressed state.

The max diameter can be anything you like.  Increasing the diameter (or the height) increases the static head (i.e. the weight of water in the header pipe) and thus the amount of water your shuttle must lift.

When discussing the shuttle volume it might be better to think of it as the capacity of the shuttle.  My shuttle never expands or compresses, rather it is the air that it contains which is compressed at the top of the cycle into the shuttle (which is simply a closed cylinder).  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.  At this point the air under the shuttle will provide lift to the head.  We can ingnore the air inside the shuttle at this point since it is the shuttle's displacement (or physical volume) being equal to its weight (because the calculator sets the weight equal to its displacement) that makes the shuttle bouyancy neutral. In fact the weight of the shuttle needs to be slightly greater than its displacement so that it will sink and the air under the shuttle needs to be slightly more than the static head so that it will 1. lift the head above the top of the header tube and 2. lift the slightly unbouyant shuttle.  I will make a note to include these factors in the calculator. I hope this all makes sense.

from your program "We know that a depth of 10m will compress the air 2 times its volume at the surface.
height / 10 + 1".  not sure if this is worded right.

I can't have worded it right or it would have been crystal clear to you.  If a balloon contains 1 litre of air at sealevel then at 10m depth it will be compressed to a volume of 0.5 litres.  I should probably have said "We know that a depth of 10m will compress the air to half of the volume it occupies at the surface."

Thanks for all the compliments guys.  I will continue working on it.

[tbird]

good morning,

i know how tempting and easy it is to take something new you just bought out of the box and try to use it instantly without reading the operating instructions only to find out you weren't as clever as we thought and then go hunting for the little piece of paper that came with your new purchase so you can make it work.  everybody is guilty of this.  i believe this is why the elsa never really got off the ground before.  the instructions (Mr. Herring's drawings) aren't as clear as they could be, but if you take the time to really study them, you will understand how it works and be able to build a working model.

maybe a rewrite of his drawings will be needed.  for sure if you don't read them, you may eventually end up with something that works but at an extra expense of time and frustation (possibly money too).

i have in previous post presented his design of various parts and methods, but seems to be like water on a duck's back.  since you insist on writting your own instructions, and sometimes physics, i will just be on the side and if i can throw in an obvious correction, i will.  good luck

tbird

[tbird]

Here is a new diagram.  I hope it helps.

Picture 1:  the shuttle is at the top and we pressurise the reservoir using valve 1.

Picture 2: the shuttle is at the top and we let the air out of the displacement area by using valve 3.  Now the shuttle descends.

Picture 3: the shuttle is at the bottom and we force the water out (displace it) by opening valve 2.  Now the shuttle ascends in picture 4.

The volume of the shuttle including the reservoir is 1.5 litres: 1 litre in the displacement area and 0.5 litres in the reservoir.  At 10m the pressure is 2 bar so the shuttle will contain 3 litres of air in the space of 1.5 litres.  When the reservoir is fully pressurised it contains 3 litres in a volume of 0.5 litres and is therefore at 6 bar.  When the shuttle is at the top the reservoir still contains 1 litre of air in a volume of 0.5 litres so it is at 2 bar.

Looking at this cycle we need to figure out two things:  how much water will the shuttle lift and how much energy will be needed to recompress the reservoir.  These I have calculated in my earlier message.  I think that model is as simple as we can get.

prajna, if this is the shuttle you plan to use, i think you have a problem.  steps 1 & 2 cancel each other.  if you put air in and then let it out so it will sink, you won't have air for step 3.  maybe you just left out another step that would leave you with enough compress air.

i thank you again for your efforts with the program.  it will be very useful, even if you have read the instructions (i think you have, more so than anyone else)

tbird.