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Author Topic: Gravity Mill - any comments to this idea?  (Read 100914 times)

ooandioo

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Re: Gravity Mill - any comments to this idea?
« Reply #30 on: August 24, 2006, 09:21:26 AM »
tbird, you are right, that is the way I'm thinking about.

hartiberlin, the inverted "U" is also a good idea.
I have calculated something: Never mind if we fill a baloon under water or a inverted "U" the pressure we need to do that depends on the water pressure+airpressure. Air pressure is 1 bar and waterpressure is 1 bar for every 10 m. Filling a baloon at 10 m under water we will need pressure more than 2 bar (29psi).
p=F/A (pressure=force/area)
F=m x a (force=mass x acceleration)

We now should see, if the amount of water thats beeing pressed upward is enough to fith these formulas, thats the crux.

Andi.

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #31 on: August 24, 2006, 02:28:09 PM »
hi stefan,

i'm not sure why you keep asking

 
Quote
But how many centimeters or inches must then the swimmer
unit be compressed at what size dimension ?

And how much energy does this use up ?

without knowing what size unit you are talking about, no answer can be given.  it's like asking how much gas does it take to drive across country without saying what transport will be used and how far across you want to go.  those answers can be figured out if we know what the size of the unit is.  this in itself needs to be determined by the user/builder.  how much work do you want to do?

Quote
Maybe someone can post a step by step calculation
of an example with all mathematical steps to calculate that,
so it it getting more clear ?

i have already done this in reply 15.  except for figuring out how much work the water we moved aboved could do.  i also said in that reply

Quote
if i made any math mistakes, please rap me on the knuckles and let me know.

why haven't i had a knuckle rapping?  are you not reading the post?  are you not following my explanations?  if not, why not ask me to re-explain where you got lost?  i can't lay the blame on you.  if you don't understand, it's only because i didn't do a good job.  for that i apologize.

the mistake i made was "now when we release the piston at 11.33 ft, it will expand to not only the orignal 1 cubic ft, but to a plus .1psi."  if you read on, you will see the figure to gain equal work was less than the pressure we increased to after figuring out the shuttle piston wouldn't expand at depth.  when i suggested the pressure to compensate for the depth (11.33ft.) should be 5psi, this was wrong.  all that was needed was 4.9psi.  then adding the amount of pressure to do the work (move water) .449psi we would end up with a 5.39psi (that was right).

the figures i used in this example were based on a size i created in the example.  this unit, for it's size would hardly produce any work.  it was a starting point.  work is not cheap.  the more work we want, the more we have to pay. to get more work from the example;  we could compress the shuttle piston farther (more work in, higher psi).  with 21.66 cubic ft of water available per cycle and/or extending the leverage of the compressing unit, we have the resource available to do extra work.

hope this clears up things a bit.  forgive me if i was too tuff on you.  btw, did you check out the link in my last (ps at the end)?

tbird

ooandioo

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Re: Gravity Mill - any comments to this idea?
« Reply #32 on: August 24, 2006, 02:38:08 PM »
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.

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

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #33 on: August 24, 2006, 02:42:14 PM »
hi stefan,

a light just came on.  are you trying to get me to WRITE a formula that you could stick in whatever variables you want and come up with answers for the unit you have in mind?  if so, you give me way too much credit.  i make too many mistakes.  they maybe simple mistakes, but the formula you seek can't have any.  it has to work for all samples.  maybe there is somebody out there watching that can.  please step up if you are willing to help.

tbird

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #34 on: August 24, 2006, 03:07:00 PM »
hi  FreeEnergy,

not sure what you are saying.  are you suggesting the shuttle piston be 6ft long or the tube the shuttle piston rides in?  also don't understand the function (purpose) of the one way valve.  can you try again?

tbird

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #35 on: August 24, 2006, 03:34:41 PM »
hi andi,

you're moving right along.  well done.

i have a question or 2.

"This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg"

"V=785.398cm? "

is the first V value just another way to write the second (never saw this before)?

"its force is F=m*g=7.704 N"  what is g?

also not sure what you mean "Looks very good, as we need quite less lifting power."  less than what?  what are you comparing it to?

"E=m*g*Dh/3.600.000J/kWh"  is Dh distance of height?

one last.  if the fall distace were .75m, does that mean power would be 1/2 again more (1.6kwh)?

tbird

ooandioo

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Re: Gravity Mill - any comments to this idea?
« Reply #36 on: August 24, 2006, 04:00:54 PM »
hi andi,

you're moving right along.  well done.

i have a question or 2.

"This will be V=7854cm?*1000cm=785.398.000cm? this weights 785.398kg"

"V=785.398cm? "

is the first V value just another way to write the second (never saw this before)?

Quote
I mean, 785.398.000cm? equals 785.398 kg water mass.

"its force is F=m*g=7.704 N"  what is g?

Quote
g=gravitational acc.=9.81m/s?

also not sure what you mean "Looks very good, as we need quite less lifting power."  less than what?  what are you comparing it to?

Quote
I'm comparing to the mass thats above sealevel, thats the mass that has to be pumped with the shuttle.

"E=m*g*Dh/3.600.000J/kWh"  is Dh distance of height?

Quote
right

one last.  if the fall distace were .75m, does that mean power would be 1/2 again more (1.6kwh)?

Quote
also right.

tbird


tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #37 on: August 24, 2006, 06:03:03 PM »
hi andi,

got most of your answers. 

still not sure what mass above sealevel you are comparing.  we are talking about your design that only works on the up stroke, right?  are you comparing the weight to produce the pressure needed to compress the shuttle piston in the first design to.... something?  i'm stuck.

help!

tbird

ooandioo

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Re: Gravity Mill - any comments to this idea?
« Reply #38 on: August 24, 2006, 06:24:26 PM »
hi tbird,

the shuttle has to be lighter than the water it displaces, then it wents up. In a closed pipe, heading 1m above sealevel, it also has to overcome the weight of the water in the pipe above sealevel, that is 785,4kg. It has to be lighter, or its volume has to be bigger than its weight in compare of the weight of the water it displaces+the water it shall displace in the pipe above the sealevel (i think i don't understand the sentence myself...).

Andi

tbird

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Re: Gravity Mill - any comments to this idea?
« Reply #39 on: August 24, 2006, 06:42:00 PM »
andi,

i think i see why we confused each other.  the water being pushed out doesn't stay above the pipe (creating back pressure). it could be delivered to a holding device supported above sealevel (you used that word) but lower than the max height the shuttle can push the water to. once the water starts to fall, work can then be extracted.  after we get water delivery (a waterfall) accomplished, then we can design the best way to make work using this falling water.  does that change your figures any?

tbird

hartiberlin

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Re: Gravity Mill - any comments to this idea?
« Reply #40 on: August 24, 2006, 07:26:49 PM »
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

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Re: Gravity Mill - any comments to this idea?
« Reply #41 on: August 24, 2006, 07:40:12 PM »
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

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Re: Gravity Mill - any comments to this idea?
« Reply #42 on: August 24, 2006, 08:32:39 PM »
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

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Re: Gravity Mill - any comments to this idea?
« Reply #43 on: August 24, 2006, 09:00:13 PM »
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.

Quote
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

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Re: Gravity Mill - any comments to this idea?
« Reply #44 on: August 24, 2006, 09:42:15 PM »
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