Gravity Mill - any comments to this idea?

[tbird]

hi stefan,

i think if i do as you suggested in your reply #40, i might be getting the cart before the horse.  if you re-read my post 25 & 27 and go to the links on those post, and take a few minutes to think about it, most of your lack of understanding what the design is, will go away.

Andi I don?t understand this part ! Did you have a typo in it ?
How do you get now more than 785 tons of water moved ??
I think there is an error somewhere and should this water go up or down

i noticed the same thing, but if you do the math, his weight per meter of water in the tube is right. (maybe i made a mistake so the answer came out right)  i think people from diffferent parts of the world use the . & , with numbers to mean different things.  i use the . for decimals and the , to seperate the larger whole numbers (every 3 places).  is that ok with everyone here?

Maybe the easiest mechanical setup would be to have piston like swimmer unit which uses internally compressed springs to have its volume size small and use a double latch magnet relay to switch/toggle electrically via short voltage pulses between bigger and smaller volume. What do you think? You could power this control circuit via an accumulator and use a waterwheel and generator to recharge the accumulator. This then should run forever and also deliver additional output energy.

this sounds like a clever arrangement, but i'm not too excited about putting electronic underwater.  maybe if i were as familar with the device as you are,i would feel different.

hi andi,

First to tbird. If the pipe reaches 1m outside the water, there is ALWAYS 785,4kg of water in the upper pipe part, needing to be moved.

i think we need to get this cleared up.  water in a pipe that is 1 meter in diameter and 1 meter long will weight 785.4 kg or 1728 pounds.  do we agree?  so if you have 10 meters of water to be moved, you will be moving 7,854kg or 17,280 pounds.  right?

The calculated energy is kWh - if you leave the /36.000.000 you have Joule. I thought about its easier to understand with kWh

i don't have any problem with this.  i just think that if it takes 2 hours to move whatever the correct amount of water is, the kwh would be different, as compared to only taking 30 minutes, wouldn't it?

hi FreeEnergy,

i'm getting the picture now.  of course the valve would lock in place at the bottom so the air can't get out.  now, this means you only have water move out on the up stroke, right?  in your 6ft long pipe, how long would the shuttle be?  remember whatever size the shuttle is will reduce the amount of water between it and the surface.  so if the shuttle is 1 ft long, it only has 2 feet of water to push.  but being only 1 foot long you can only push up 1 foot if the pipe remains the same size above water.  if the shuttle is 1.5 feet long, you will have 1.5 feet of water to push up and it will rise the majority of the 3 feet out of the water.  it will be shy by the amount the shuttle displaces at the surface.

do you figure to fill it with a compressor like andi plans?

maybe in the future we can refer to the type unit we will be talking about when we start.  i suggest using type 1 for the double action unit and type 2 for the single action type.  what do you think?

[ooandioo]

Hi tbird.

Right, I'm using "." as seperator and "," for decimals - sorry for that.

i think we need to get this cleared up.  water in a pipe that is 1 meter in diameter and 1 meter long will weight 785.4 kg or 1728 pounds.  do we agree?  so if you have 10 meters of water to be moved, you will be moving 7,854kg or 17,280 pounds.  right?

Right, we are moving 7,854kg, but we do only lift the weight of water thats in the upper 1m (785kg), whatever the way from the bottom to the top is, if not - there would be no overunity.

i don't have any problem with this.  i just think that if it takes 2 hours to move whatever the correct amount of water is, the kwh would be different, as compared to only taking 30 minutes, wouldn't it?

The unit kWh only shows what is possible - surely we have to divide it by the time we will need to lift the water up. E only shows what amount of energy lies in the lifted water at all.

Andi

[hartiberlin]

Okay Andi and TBird lets try again step by step to calculate it,
if we can get more potential energy from moving the water into an upper
reservoir: potential-Energy= watermass x earthaccel. g  x heightdifference,
than needing for the compression of the swimmer unit !
Okay, let?s stay with Andi?s example:

A pipe 10 Meter deep inside the seawater and 1 Meter above the seawater level.
diameter = 1 Meter


Now let us just use the lifting not the sinking to produce the energy.

Okay, now the START condition.
All water in the pipe is in the seawaterlevel position and the
piston like swimmer body must be 2 Meters under seawater level
to be able to go up to seawaterlevel and thus
pump 1 Meter of volume over the 1 Meter high pipe into an upper reservoir.

Okay, how big will be the potential energy we win, if we succeed to pump this
volume of 1 Meter height of 1 Meter diameter water up:

Volume=Area x height, Area=r?*Pi=2500cm*Pi,
V=785398cm? this means the water weights 785.4kg

Thus we have a potential energy stored of:

Energy= 785.4 Kg x 9.81 x 1 Meter heightdifference= 7705 Joule or Wattseconds.
That means put into Watthours we must divide it through 3600, so
we get about 2.14 Watthours of potential energy stored 1 Meter above sealevel with
about 785.4 kg= 785.4 Liter of water !


Now let us see, if this would be enough energy to compress the piston
swimmer body again, so he can go down again 1 Meter now under seawaterlevel.


As we needed 2 Meter volume of water= 2 x 785.4 Kg = 1570.8 Kg watermass to be pushed up 1 Meter
ontop of the pistion swimmer unit,
we needed a force of F= watermass x g to push it up , that is
about 15410 Newton of force.


That means, the lifting power or positive boyuancy has to be greater then 15410 Newton.

Now how big and what volume must the piston like swimmer boddy have ?

diameter= 1 Meter
What height ?

As Archimedes's principle says:

The buoyant force is equal to the weight of the displaced fluid,
we must use very light piston swimmer body and if we do not count in its weight
we must have then also 2 Meters of height of the piston swimmer body, because it must have the
same force as the Mass force of the 2 Meter water volume above it.
So to get on the positive site, the swimmer body must be a little over 2 Meters high himself
also depending on its own weight.


Okay, now at the starting point, the swimmer unit is 2 Meters under seawaterlevel
with its upper surface and 4 Meters under seawaterlevel
with its lower surface.
When it has risen up the 2 Meters , 1 Meter of volume of water is pushed up into the
upper reservoir and the swimmer is with its upper surface at seawaterlevel
and ontop of it we still have a volume of 1 Meter water, which now
weights still 785,4kg, which is a force of about 7705 Newton onto the upper surface
of the swimmer unit.
The swimmer unit gets from the buttom a buyoancy force of 2 x 7705= 15410 Newton,
so we have to reduce the positive buyoancy to lower than 7705 Newton to get the swimmer
body again to sink.
This can only be done by compressing it to a little bit less than  1 Meter height, so the
1 Meter height water above it has enough weight to sink it down again.

Now let us calculate the pressure inside the swimmer body, which we assume will be a balloon
for the moment.
At the starting point 2 Meters under seawater level the 2 Meter high balloon must have an airpressure of
P= F / A of  P=  15410 Newton / 7853.98 squarecm =1.96 bar
( that is about the pressure in my car wheel ! for comparison only)

So this is the pressure to keep the volume of 2 Meters water above the ballon inside the 1 Meter diameter pipe,
so that the balloon will not be compressed and still has its shape of 2 Meter height itsself.

Now in the end position, when the upper balloon surface area is at seawaterlevel and only 1 Meter
of  Watervolume is still ontop of it, the ballon will have a pressure of:
P= F / A of  P=  7705 Newton / 7853.98 squarecm =0.98  bar

So now we must compress the volume of the  balloon from 2 meter height to 1 Meter height,
this means we have to press the upper surface of the balloon 1 Meter down, so the pressure
inside the ballon rises again from 0.98 bar to 1.96 bar.
So we must apply 7705 Newton at the start to 15410 Newton at the end of the compression phase of 1 Meter
distance to do that .
But how much energy is this ?

If we do it graphically we see, that the area under Force over distance is a triangle and the
energy area under it is half the value of the full area, that the endposition would give.
So the energy required is:
Compression-Energy=15410 Newton x 1 Meter / 2 = 7705 Newtonmeter = 7705 Wattseconds = 2.14 Watthours !


So we need the same energy to compress the balloon cylinder as we have won in the first place !
So as we also have frictional and seal losses, this does not work !

Also due the hydrostatic paradoxon we will also not be able to pump
air under water into any balloon or other container, as we have to overcome
the full height of the water inside the hose, so this also needs too much energy
to pump air under water !
So this gravity mill just does not work... otherwise it would have been used probably much
earlier...

Too bad.
Regards, Stefan.

[hartiberlin]

Hmm, I am just asking myself, what happens, when the balloon has 2 Meter height
and 1.96 bar at the starting point 2 Meters under water and then rises 2 Meter and
has 0.96 bar at the end.
Then its volume must have also doubled , right ?
So must it now be 4 Meter in height ?
I am a bit confused now... Hmm...

I guess one can not just work with a balloon, that can expand
and compress itsself due to pressure !

One must use a swimmer body, that can be put into 2 volume positions
via electric relays. If you use a vaccuum inside the swimmer body,
it will be easy to recompress the swimmer from 2 Meters volume height
to 1 Meter volume height.
But then it is the question, how much energy you need to
restretch the volume of the swimmer body again from
1 Meter to 2 Meters when it has sunken again...?

Probably also more as you have won to move the 1 Meter volume water 1 Meter up !???

[tbird]

hi stefan,

nice try, but no cigar.  must be frustrating to go thru all those calcs and still come up wrong.

most of the setup you used is for a unit that travels to 1 atomsphere (10 meters) and not just 2 (or 4 as you threw in) meters.  maybe you have come up with type 3.

if you only go down 1/5th of an atmosphere, you only need to compress to .2 bar (not use to these measurements).  so everything you based on .98 and 1.96 is wrong.

there are lots of holes in your summary.  it would take awhile, but if you want me to, i will try to point them all out.

i will have to say, i probably could not have done a summary nearly as well as you did yours.

let me know about the corrections.

tbird