Gravity Mill - any comments to this idea?

[hartiberlin]

Hi TBird, okay, I have to reread your message 15 again, but am online via PDA now posting this..Maybe you can post again a summary with all steps of Mathematics of a model setup you are sure will work and also tell all dimension of pipes and swimmer, so we can see, that it works.. many thanks.

[tbird]

andi,

just converted your numbers to ones i understand.  i think you have an error in the kwh figure.  before i start barking up the wrong tree, let me make sure of a couple of things.  kwh does stand for kilowatts per hour, right?  if a hour is the time frame for power, what is the time frame for delivering 785.398kg of water.  do we know how long it will take to collect this much?  oh heck!  i can't keep it to myself.  unless you were going to limit the shuttle to rise only 1 meter, you will get almost 10 meters worth of water (7,853.98kg) per cycle.  it won't be that much because the shuttle can't travel the full meter of tube that is out of the water.  so you might only get 2/3rd of that last meter.

maybe we can't figure kwh until we know how much water will be available per hour.

tbird

[hartiberlin]

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.

[hartiberlin]

Here some more calulations in this:
Lets assume you have pipe, 10m under water and 1m over water. Its diameter is 1m.

Our shuttle has to only lift the amount of water thats in 1m above the water, thats:
V=A*h, A=r?*Pi=2500cm*Pi, V=785.398cm? this weights 785,4kg, its force is F=m*g=7.704 N
That means, the lifting power has to be greater then 7.704 N.

Okay, so far okay with me.

If this is the fact, the shuttle will push the water thats in the pipe under water 1m above the water.
This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg, its force is F=7.704.754 N.
Looks very good, as we need quite less lifting power.

One more, the pumped water has its energy:
E=m*g*Dh/3.600.000J/kWh
E=785.398kg*9,81*0,5m/3.600.000J/kWh=1,07kWh
The maximum energy we could take out of the water 1m above sealevel is 1,07kWh (if we alow 0,5m fall down).

Andi

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

[ooandioo]

Hi all.

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

hartiberlin - the idea with spring compression is worth thinking about it.
785 tons of water are in the pipe part under water.

Andi