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Author Topic: communicating Barrels  (Read 11927 times)

pinobot

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communicating Barrels
« on: January 17, 2006, 10:00:40 AM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels.gif)
I'm a bit bored.  ;D

FreeEnergy

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Re: communicating Barrels
« Reply #1 on: January 17, 2006, 12:46:20 PM »
 ::)

pinobot

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Re: communicating Barrels
« Reply #2 on: January 17, 2006, 01:12:29 PM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels2.gif)
An explanation:
The barrels are airtight, through hose X water is tapped thereby creating underpressure in A and overpressure in B that push/sucks water through Y to barrel A.
 :D

hartiberlin

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Re: communicating Barrels
« Reply #3 on: January 26, 2006, 12:31:21 AM »
Hmm, I guess it will come to a stillstand after some time and
equalice itself into equilibrium... right ?

FreeEnergy

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Re: communicating Barrels
« Reply #4 on: January 30, 2006, 11:17:38 PM »
build one and let us know

pinobot

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Re: communicating Barrels
« Reply #5 on: January 31, 2006, 01:48:44 AM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrelsfix1.gif)

small "fix".

pinobot

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Re: communicating Barrels
« Reply #6 on: January 31, 2006, 10:16:42 AM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrelsvariation.gif)

Some variations, with only one airtight barrel.

pinobot

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Re: communicating Barrels
« Reply #7 on: January 31, 2006, 11:31:09 AM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrelsexplanation.gif)

Simple explanation:
Barrel A is airtight, barrel B is not.
Gravity is used to pull a fluid from A to B through the big hose, when the fluid flows a vacuum is created in "a" through te mass of the fluid in the piece of pipe "c", if the fluid doesn't flow then only "b".
"d" is the normal level in the small hose, "e" is the hight the fluid has to be raised.The weight of the fluid to be raised in the small pipe "e" is less than the weight "c" creating the vacuum "a".
So it could work.
« Last Edit: January 31, 2006, 06:23:51 PM by pinobot »

FreeEnergy

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Re: communicating Barrels
« Reply #8 on: January 31, 2006, 12:06:32 PM »
and when you do build one post videos :)

pinobot

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Re: communicating Barrels
« Reply #9 on: February 01, 2006, 11:43:51 AM »
I found a device that works almost in the same manner designed 400 years ago on Donald E. Simanek's website. :P
http://www.lhup.edu/~dsimanek/museum/unwork.htm#sinclair

It's the George Sinclair's Siphon.
(http://www.lhup.edu/~dsimanek/museum/sinclair.gif)

It's not as good as mine. ;D

IcyBlue

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Re: communicating Barrels
« Reply #10 on: February 01, 2006, 12:10:06 PM »
Just some keywords: hydrostatic pressure, hydraulic systems => 1st semester of classical physics.

pinobot

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Re: communicating Barrels
« Reply #11 on: February 01, 2006, 12:49:47 PM »
help me out here, i know about hydrostatic pressure, what has it got to do with this design?

I know these designs don't work because of hydrostatic pressure, or will they?? :P

(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels6.gif)

(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels7.gif)
« Last Edit: February 01, 2006, 02:18:24 PM by pinobot »

FreeEnergy

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Re: communicating Barrels
« Reply #12 on: February 01, 2006, 02:14:22 PM »
more details on this one please.. :)

pinobot

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Re: communicating Barrels
« Reply #13 on: February 01, 2006, 02:40:38 PM »
(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels6explanation.gif)
Part "B" can move up and down.
The weight of the fluid "b" pulls down part "B" creating underpressure in "A", sucking up fluid through pipe "a" to "b" in the direction of the arrow.


(http://img.photobucket.com/albums/v156/pinobot/communicatingbarrels7explanation.gif)
Part "B"can move up and down.
The weight of fluid "b"pulls down Part "B" creating underpressure in "A", sucking fluid throught pip "a"to "b" in the direction of the arrow.

IcyBlue

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Re: communicating Barrels
« Reply #14 on: February 01, 2006, 05:19:53 PM »
You have a closed system. This means, if you reduce the pressure in the chamber above the fluid, you excert a dragging force onto the fluid. Since it is a closed system, we can ignore (to simplify the explanation) the fluid in the volume with the bigger diameter. What counts is only the diameter of the ingoing and outgoing pipe. Since they both have the same diameter, the underpressure you create is causing the same force on both of them - but in the opposite direction. Both forces cancel each other out, leaving you with a net force of zero. The only thing that will happen is that the fluid at some level of underpressure starts to change its state, i.e. it "boils".

The hydrostatic pressure (F=rho*g*h) on the left side (with the large fluid volume) is exactly the same as on the right side (with the way smaller volume). Only the height of the fluid column counts, not its mass. Since you connect them in a U-shape fashion, they both cancel each other out, leaving you again with a net force of zero.

The only chance to get anything in motion is to disturb the energy balance on either side of this setup, e.g. changing the gravity that acts on the fluid on one side.