Gravity Mill - any comments to this idea?

[tbird]

hi prajna,

thank you for your kind words, but they are not necessary.  your actions are quite satisfying!

now for the reason i'm posting these days, obvious mistakes.

you didn't state (anywhere that i saw) which pressure you were using, gage or absolute.  looking at your numbers,

So the pressure inside our shuttle will be 4.41  bar

i assume absolute.

after coming up with that answer, you then say

So we will have to compress 4  times our displacement volume into the shuttle i.e. 43.28 litres
[2 * compression * displacement volume]

So the pressure inside our shuttle will be 4.41  bar
[total compressed volume / shuttle volume]

So we need a weight of 432.8  kg times our compression ratio of 3:1 to recompress the shuttle.

hmmm.....  is this obvious?  i didn't do the math, is this just a typing mistake?  if not i'll explain in next post why 3:1 is wrong.

i have to say, just because i think you guys are not really working on the elsa, i will try not to be a negative.  in fact, i will help you get the device and numbers as accurate as possible so when you do get around to working on it, you will see the real differences.

this may help;  in the default example of prajna's program, he uses

"First we need to decide to what height we will pump the water:
1 (meters)

Then choose the internal diameter of our header tube:
3.57 (centimeters)"

i don't think 1 meter is the max height the shuttle can lift the water.

We know that the shuttle volume is 9.82 litres
[Ïâ,¬ r2 * height]

Therefore our shuttle will have to weigh at least 9.82 kg in order to sink since it displaces that weight of water

if you reduce this diameter, you can gain height (i believe that is part of the formula you used to determine stored energy).

also,

So we need a weight of 432.8  kg times our compression ratio of 3:1 to recompress the shuttle.
which equals 1298.4  kg

Subtracting that from the total volume of water pumped gives us an excess of 654.28  litres stored at 1 m after we have recompressed the shuttle.

i'm not trying to point out the ratio error, but the fact there is no mention of how you used the compression weight.  would it be possible to use that energy (weight) again?  maybe the water has to travel the full distance to develope the energy.  this is not my best subject.  but,... if you did use leverage (hydraulics is leverage as well as a pole) of 10 feet, you would only need 1/10th the weight.  if you put 90% of the weight you figured to use back in the excess column, how would that look to you?

one last thought.  what if you could get twice the water stored for the same compression energy?

fyi, there is no value for the kwh you converted to from the kilojoules.

tbird.

[hartiberlin]

@wizkycho
nice idea !

@pranja
how do you calculate the output power now ?
Via the gas laws or via the mechanical lever pressure ?
The output seems to be pretty low now...

[hartiberlin]

Also we need to check how high we really can pump the
water through the small exit tube with just a 2 or 5 cm high shuttle...

I did not yet have time to get better tubes and a shuttle that fits in tightly.

Maybe 2tiger with the "boxing of a Barbiefamily"  can check into it ?
Many thanks.

[prajna]

Well, tbird, it looks like I don't handle pressure well.   Going back to the 10cm x 10cm x 10cm box we were discussing in an earlier message... it has a capacity of 1 litre.  If I compress 3 more litres of air into it then it now contains 4 litres of air.  What does the pressure guage read?

As far as the shuttle is concerned we need to take the original 1 litre of air into account because it is a factor in the calculation of how much water will be displaced at 10m depth.  The total 4 litres will expand to displace 2 litres of water.  Now, suppose I used gauge pressure in my calculation: the calculation would be wrong.  Balanced against that we have the problem that for anyone who frequently works with gauge pressure, they will look at the figure of 4.41 bar and think 'Man, that can't be right.'  Which is your reaction, understandably.  For anyone who is not so familiar with the difference between gauge pressure and absolute pressure it is more intuitive (I think) to comprehend the calculations if they are expressed as absolute pressure.  Perhaps I am wrong or perhaps I should just indicate that the pressure is absolute.

Where I calculated the pressure I have included the formular I used:

So the pressure inside our shuttle will be 4.41  bar
[total compressed volume / shuttle volume]

So anyone who is concerned should be able to see what I did.

As far as the compression ratio is concerned, how should I calculate it.  I am not sure why I chose to calculate it this way.  Perhaps you can help here.

As to ELSA, I haven't studied it.  I did take a look at John's pages but only sufficiently to get a rough idea of what he was up to and did all the rest from 1st principles (which, as I explained to you in an earlier post, is my preferred way of working.)  If this system is wildly different than ELSA then perhaps we could refer to it as 'ELSA inspired.'

The default value of 1m for the header pipe was just an example.  You have probably already discovered that you can set it to whatever you like.  1m by 3.57cm gives a head of 1 litre and that is the only reason I set those values as defaults.  My preferred height is 5m.

By the way, in case you haven't already figured out, all of the values that are in bold type in the text are calculated (or copied from calculations further up the page).  Perhaps I should point that out on the page too.

if you reduce this diameter, you can gain height

Yes, of course.  You can change any of the values from their defaults and then click the 'calculate' buttons to recalculate the results.  That is the whole point of the application.

'm not trying to point out the ratio error, but the fact there is no mention of how you used the compression weight.  would it be possible to use that energy (weight) again?  maybe the water has to travel the full distance to develope the energy.

Certainly we can compress more efficiently by using some fraction of the water we currently use for compressing via a lever.  I am building this thing up bit by bit, as you have seen, and I wanted to start with the simple case.  We can add a calculation for that as well.  It will simply require adding an input box for the lever-arm lenght and the appropriate calculations.

one last thought.  what if you could get twice the water stored for the same compression energy?

It would look very much like the lever system we have just discussed.

fyi, there is no value for the kwh you converted to from the kilojoules.

There is on my screen:

That might sound like a lot but it represents a stored potential energy of 6.41  kilojoules  which is approximately 0.002  kwh

It should display as 0 if it works out at less than 0.0009 kwh.

Thanks again for your careful evaluation, tbird.

@stefan
I use the mechanical pressure (mgh)

[tbird]

prajna,

i'm going to start a little out of order.

Balanced against that we have the problem that for anyone who frequently works with gauge pressure, they will look at the figure of 4.41 bar and think 'Man, that can't be right.'  Which is your reaction, understandably.

you misunderstood my meaning.  i do think the figure is right.  it's the ratio that bothers me.  maybe i'm wrong, but i have always been under the impression if you have something that is 4 times something, then the ratio is 4 to 1, not 3 to 1, right?  have i lived under the wrong impression all my life?

Now, suppose I used gauge pressure in my calculation: the calculation would be wrong.

the first thing i want to say is why?  i guess i know why, but can't the calculation be rewritten?  the only thing you would have to do is divide by 2 at the end.  in your example the 4.41 would become 2.205, which would be right.  you could kinda compare it to why we don't have to worry about the water in the pipe.  it all takes care of itself.  you will always need a greater difference inside the container than it is in to get it to expand.  since we are always (maybe we shouldn't) want to double or halve our volume, simply determine the gage pressure at the depth (this will make it static) you want, then to expand just double that amount.  works every time.

For anyone who is not so familiar with the difference between gauge pressure and absolute pressure it is more intuitive (I think) to comprehend the calculations if they are expressed as absolute pressure.  Perhaps I am wrong or perhaps I should just indicate that the pressure is absolute.

the best way i can commit is comparing metric to what i grew up with (and visa versa for stefan).  i really struggle getting thru the math in metric, not to mention to visualize without converting is impossible.  if you have never been exposed to a gage (never checked your tire pressure, never been scuba diving, etc.) then it absolute will be your standard.  you can see where i'm going.  gage is easiest for me (and i know about absolute).  one more question you can ask yourself.  why isn't absolute pressure used more?

The default value of 1m for the header pipe was just an example.  You have probably already discovered that you can set it to whatever you like.  1m by 3.57cm gives a head of 1 litre and that is the only reason I set those values as defaults.  My preferred height is 5m.

this comes back to one of my orignal suggestions (quack quack).  if you build a unit, what do you know?  not much.  the fewer details you have to provide, the better for the user.  the first thing you might know is the size of the pipe available.  you might have (could be required) a size of the shuttle in mind to be used.  remember, the first brick hasn't been laid (so to say).  why not give the user the max height, using as a standard 10% of shuttle pipe size, or something reasonable, automaticly.  this would be better than guessing about the height.  the user could still enter other sizes, but i think most people would rather see the highest energy practical first.  don't get me wrong, if you left it as is, it would still be a wonderful tool.

It would look very much like the lever system we have just discussed.

no. it would only have to use both strokes.  water delivery just doubled.  if time is a factor (we haven't ever addressed that), then we need to find out which one does the best job.  if not, twice is better with only one recompression.

speaking of time.  when you figure energy stored, simply raising the delivery height increases the energy stored.  wouldn't time have to be considered somewhere?  if we have a working unit and decided to raise the tank twice as high, this would take twice as long (if shuttle stayed the same) to deliver the same amount of water.  soooo.....  am i missing something?  any thoughts?

enough for now.

tbird