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Author Topic: Self Siphoning Water  (Read 35967 times)

Offline tbird

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Re: Self Siphoning Water
« Reply #30 on: August 22, 2010, 08:38:01 PM »
hi guys,

i have a question.  i know you said before, the small tube on the right should enter above the water in the tank, but should the exit tube on the left have its entrance (top at the tank) above water?  is that just an art mistake and you really mean for it to be exposed to the water in the tank?  and those 2 thin black lines are just left over art that mean nothing?

point of interest...the left tube now looks like my old chicken waterer, if the top were all closed up.  water would stay inside until the outside level was lower than the lip of the pipe (we used a jar).  then air would go in and water would refill the outer area, covering the lip again.

tom

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Re: Self Siphoning Water
« Reply #30 on: August 22, 2010, 08:38:01 PM »

Offline tbird

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Re: Self Siphoning Water
« Reply #31 on: August 22, 2010, 09:50:33 PM »

i don't think you understand my question.  must be my accent.  let me put it a different way.

the tube on the left has an elbow at the top that looks like it is connected to the tank just above tank water level.  also the elbow shows to have air in it, which would be logical since it enters the tank from the side above the water in the tank.  my question was, is that the way you meant for us to understand the picture or is that an art (drawing) error?

the same question for those 2 small lines, one on each pipe close to where they go to the tank.  are they just left over drawings that mean nothing?  or are they some kind of valve?

tom

Offline Low-Q

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Re: Self Siphoning Water
« Reply #32 on: August 23, 2010, 04:14:36 PM »
  @aLL,
 This is to explain basic hydraulic theory. In the design we're discussing,
vacuum is replacing pistons, hopefully. It might be an untested idea  ;D

 With the attached pic, if A is PiR^2 = 3.14 x 2^2 =
                                                      3.14 x 4 = 12.56^2 (square) inches

 and if B is P x 1^2 =
                  3.14 x 1  = 3.14

 What hapens is the pressure applied to A is transmitted to B increasing the potential.
An example is if 5psi of force is applied to A, then it is 5 x 12.56 = 62.8

 Since B is 3.14^2 inches, we divide 62.8/3.14 (B) = 20 psi
The reason this happens is the piston focuses the force on the shaft connecting the two pistons. And the resulting pressure is the total pressure applied to piston A divided by piston B surface area.
 It's simlar to gear ratios in transwmissions and drive axles/differentials on cars.
 So what happens is because piston A has 4 times the surface area of piston B, the pressure is amplified 4 times.

 
                                                                                         Jim
Yes, but the volume are also reduced to 1/4... So 4 times the pressure does not help if the volume of the pressure are 1/4. Then you end up with the same potential energy in both volumes. No difference in potential energy will not be possible to harness.

Vidar

Free Energy | searching for free energy and discussing free energy

Re: Self Siphoning Water
« Reply #32 on: August 23, 2010, 04:14:36 PM »
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Offline FreeEnergy

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Re: Self Siphoning Water
« Reply #33 on: August 23, 2010, 09:32:33 PM »
See the Open Source Vs. Patenting topic over here: http://www.overunity.com/index.php?topic=1821.0

Offline ResinRat2

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Re: Self Siphoning Water
« Reply #34 on: August 24, 2010, 12:50:58 AM »
Hello Gentlemen,

I have not been commenting because I feel I am missing some detail, forgive me if this has already been addressed, but I can't see how you can have a vacuum and an opening to the atmosphere in the same system.

Nature hates a vacuum, so either air or water will fill it. Wouldn't the water just run out and the area at the top start to fill with air or water? Wouldn't the water just run out on the left (picture) side? What would you think would stop it from happening?

Again, sorry if I misunderstand, but I am certainly missing something vital here.


Free Energy | searching for free energy and discussing free energy

Re: Self Siphoning Water
« Reply #34 on: August 24, 2010, 12:50:58 AM »
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Offline tbird

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Re: Self Siphoning Water
« Reply #35 on: August 24, 2010, 02:27:10 PM »
Quote
This is to explain basic hydraulic theory. In the design we're discussing,
vacuum is replacing pistons, hopefully. It might be an untested idea 

 With the attached pic, if A is PiR^2 = 3.14 x 2^2 =
                                                      3.14 x 4 = 12.56^2 (square) inches

 and if B is P x 1^2 =
                  3.14 x 1  = 3.14

 What hapens is the pressure applied to A is transmitted to B increasing the potential.
An example is if 5psi of force is applied to A, then it is 5 x 12.56 = 62.8

 Since B is 3.14^2 inches, we divide 62.8/3.14 (B) = 20 psi
The reason this happens is the piston focuses the force on the shaft connecting the two pistons. And the resulting pressure is the total pressure applied to piston A divided by piston B surface area.
 It's simlar to gear ratios in transwmissions and drive axles/differentials on cars.
 So what happens is because piston A has 4 times the surface area of piston B, the pressure is amplified 4 times.

hi jim,

i think this is where your theory is flawed.

using the pistons, you are right about the increase in psi to the smaller piston.  but what happens if the pistons are replaced with water in a tube?  how much pressure is now applied to the smaller container?  since there is no longer a collection device (the piston), the only force the smaller side sees is the head pressure (psi) created by the height of the water column in the bigger tank.

hydrostatic paradox is what we run into.  have a look

here..http://scubageek.com/articles/wwwparad.html

you may say why doesn't it work like a piston.  it kinda does.  all the pressure, except for the opening to the small tank, is applied to a fixed object, the container walls and bottom.  if they could move, you could use the benefit of the extra weight.  since they don't, all you get is the small amount applied thru the small tube.

take your drawing and make it a container of water and open the top of the big side.  next to the connecting path (between the 2 tanks) on the big tank, drill a hole.  note the force of the water coming out.  now (with the same amount of water in the system) drill a hole, at the same height, on the opposite side of the connecting path in the small tank.  according to your theory, the water coming out should be 4 times greater.  i think you know it won't be.

i hope you see now what i've been trying to say.

tom

Offline tbird

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Re: Self Siphoning Water
« Reply #36 on: August 24, 2010, 04:03:35 PM »
i understand testing, but if i see a hole in my bucket, do i still fill it with water to see if it will leak?

Quote
As such, the person that started this thread believes people should give their idea's to businesses for free. All they have to do is make some claim and call it something else or cite a different source for the idea and it becomes new and original Why open sourcing won't work. People don't build which would be a requirement of open sourcing giving power to the people

are you trying to make a point here?  if so, i don't get it.

tom

Free Energy | searching for free energy and discussing free energy

Re: Self Siphoning Water
« Reply #36 on: August 24, 2010, 04:03:35 PM »
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Offline ResinRat2

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Re: Self Siphoning Water
« Reply #37 on: August 24, 2010, 04:12:04 PM »
  RR2,
 I went to school for this. Could explain but will leave it to FreeEnergy.


Never mind. I should have just remained silent. Please ignore my last posting. Thanks.

Offline FreeEnergy

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Re: Self Siphoning Water
« Reply #38 on: August 24, 2010, 07:34:05 PM »
i started this thread with the name "Self Siphoning Water" because i also use siphoning in the system.
i wasn't really thinking much to give it a name, just something i quickly thought of.
probably after seeing your thread this name was in the back of my head and i subconsciously used it.
if you want you can ask the Moderator to change the name or delete this thread for you if it bothers you that much, i give you permission lol 

as for the "Self Siphoning Water" system ideas, i don't know if it will work! probably wont work. but you never know ;-)


Open Source is the only way to go, at least for me and many others.
Open Source can be very successful in the business world too, Linux is just one example for starters.
Anyway this is not the place to talk about this, we'll leave this for the Open Source Vs. Patenting thread.

well i think i am done with this thread.

Free Energy | searching for free energy and discussing free energy

Re: Self Siphoning Water
« Reply #38 on: August 24, 2010, 07:34:05 PM »
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Offline Low-Q

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Re: Self Siphoning Water
« Reply #39 on: August 27, 2010, 11:28:16 PM »
I have a question (Without having read all the posts):
If you have  bucket of water, and make a siphon. Let's say we have a hose in the water and out of the bucket, pointing towards the gound so the siphon effect occour. In the bottom of the hose this hose is connected to an "Y"-split, and slightly spiral shaped so it can rotate as the water pressure goes through the ends. Lets say it rotates the opposite direction of the earth so we can harness this effect too (Do not remember the name of it).
My guess is that this rotation will use a centrifugal force that will slightly increase the siphon effect so the bucket of water can be emty faster. This means the kinetic energy of the rotating water exceeds the potential difference between the water level in the bucket and the rotating end.
So let's further make a device that directs this water upwards towards the bucket so we can refill it with the same water.

Is this something to think about?

Vidar

Offline Low-Q

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Re: Self Siphoning Water
« Reply #40 on: August 29, 2010, 12:08:33 AM »
I have a question (Without having read all the posts):
If you have  bucket of water, and make a siphon. Let's say we have a hose in the water and out of the bucket, pointing towards the gound so the siphon effect occour. In the bottom of the hose this hose is connected to an "Y"-split, and slightly spiral shaped so it can rotate as the water pressure goes through the ends. Lets say it rotates the opposite direction of the earth so we can harness this effect too (Do not remember the name of it).
My guess is that this rotation will use a centrifugal force that will slightly increase the siphon effect so the bucket of water can be emty faster. This means the kinetic energy of the rotating water exceeds the potential difference between the water level in the bucket and the rotating end.
So let's further make a device that directs this water upwards towards the bucket so we can refill it with the same water.

Is this something to think about?

Vidar
Anyone?

Free Energy | searching for free energy and discussing free energy

Re: Self Siphoning Water
« Reply #40 on: August 29, 2010, 12:08:33 AM »
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Offline fritznien

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Re: Self Siphoning Water
« Reply #41 on: August 29, 2010, 04:27:59 AM »
whats to say he is talking about a pump.
fritznien

Offline Low-Q

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Re: Self Siphoning Water
« Reply #42 on: August 29, 2010, 06:33:55 PM »
whats to say he is talking about a pump.
fritznien
That is what I am talking about too. However I was thinking if it was possible to use the rotation of the earth to enhance the rotation, and force the water in the end of the hose faster so the centrifugal force could be enough to squirt that water up in the bucket for reuse.

Vidar

 

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