I was messing around drawing and came up with this idea,
If you've studied gravity, water siphoning, and capillary water tubes, you should see a clear picture.
the drawing was done free hand by me so its not to its exact measurements.
- edit -- edit -- edit -- edit -
the water tube should be more like a quarter of a circle shaped type of tube.
and i guess the water reservoir should hold a lot of water, probably more than the tube itself.
but the water would have to be spread out in low depth?
and the hoses should be equal in diameter?
- edit -- edit -- edit -- edit -- edit -
MADE A SMALL CHANGE IN THE DRAWING PLEASE REVIEW.
http://demo.physics.uiuc.edu/lectdemo/scripts/demo_descript.idc?demoid=230
thanks for that.
this is pretty easy to test out.
anyone else think this will work?
http://www.lhup.edu/~dsimanek/museum/unwork.htm#capillary1
Quote from: P-Motion on August 13, 2010, 09:49:24 AM
ResinRat,
The idea is not based on capillary action. I can go into more detail and explain the difference between capillary action, hydraulic theory and how using vacuum is different.
Jim
Sorry, I misunderstood
RR2
As is this concept looks flawed considering atmospheric pressure. However you made me thing about an alternative idea. Will post later.
broli,
please post asap!
:-)
Ok as promised.
Your idea made me think of using 2 concepts to create 1 result. It uses siphoning action and the fact that a high enough water column can overcome athmospheric pressure. However the big IF is whether indeed a siphon action will take place and cause a vacuum at the top of the sealed container instead of break down the water continuity and cause a vacuum at the top of the tubes.
Siphons in vacuum work but combining that with a water column high enough to overcome atmospheric pressure is new to me. If it indeed creates a vacuum at the top of the sealed container then you can guess yourself what benefits that has.
hi guys,
i've been watching this thread but still can't figure out what you are trying to accomplish. are you trying to get a continuous flow of water to drive a power source?
tom
This is how I picture it working.. Just with a little air pressure and one way valves. Air tight container.
Sorry, the water will not flow no matter how much pressure you pump into the container. The air pressure pushes equally in all directions and will not allow flow into the open container space through the one way valve.
Well water cannot be compressed and air can... I guess I was thinking optimistically when I thought the specific gravity of the water would be able to over power the air pressure.
Quote from: P-Motion on August 15, 2010, 10:38:35 AM
@Free Energy,
Here is what I came up with. I decide ed to try doing it on my paint program.
What I like about having a reservoir on top is the "cool" factor. Everybody is used to seeing water drain at the bottom of a pipe, but at the top ?
I think that is something that would catch people's attention. Of course, the larger the volume of the top reservoir, the more vacuum that would be needed. it's shape could be modified to lessen it's volume.
Of course, like anything else, to see if it works would take some careful work to have everything set up properly. Then by knowing all the different values, it could be properly tested.
edited to add; and of course, the water level in the top resrvoir would always stay at the same level !!!
end edit
Jim
p.s. in my previous post with the 2 cylinders, the 5 pounds of water only creates 1 psi of downward pressure if the surface area of the tube is 5^2". I said it is still only 5 psi the second time I mentioned it. I should have said 1 psi.
in your drawing i don't think the water reservoir will hold the same level of water during its process, the water would run out and air bubbles would go up the tube and it would lose vacuum in the reservoir section. the system would stop. this is why i would use equal diameter tubes. i dont even think a vacuum is necessary, it might help or not gotta try it out.
-edit-
need to try my original idea in this thread!
Quote from: tbird on August 15, 2010, 12:22:36 PM
hi guys,
i've been watching this thread but still can't figure out what you are trying to accomplish. are you trying to get a continuous flow of water to drive a power source?
tom
um, no that's not it at all, complete opposite of that!
are you here tell us these kind of ideas will NEVER work? lol
Quote from: P-Motion on August 15, 2010, 04:08:50 PM
Tom,
Initially, just to see if a continuous flow can be achieved. If so, then that alone would be something.
And since it would have a drain, it would be possible to power a water wheel. Wouldn't generate much energy, but could generate interest in alternative engineering.
@All,
Had a thought about filling the system. What is interesting is by back filling it, by putting a tube oover the discharge and creating a greater static head, water could flow thrrough the drain into the resrvoir on top and then into the inlet pipe.
And then, when the fill tube is removed, it would either work or not work. I think the reservoir being filled to the level of the inlet pipe would help it to flow quicker if it were to work. A small detail that I think is important to remember is that the reservoir could only be filled to just below the level of the inlet pipe to maintain seperation between inlet and discharge sides of the system.
@Free Energy, when the drain tube curves up, it creates a trap that won't let air in. Under your sink, it is called a loop seal that keeps sewer gasses from flowing out of your sink.
edited to add; This might also help some to understand how I think Bessler's wheel worked. I do believe he used water.
which way is the water flowing?
--edit--
never mind, i see it now!
when the water reservoir fills up to a certain water level the water gets flushed!
Quotebroli
Hero Member
Posts: 1588
Re: Self Siphoning Water
« Reply #8 on: August 14, 2010, 09:23:02 AM »QuoteOk as promised.
Your idea made me think of using 2 concepts to create 1 result. It uses siphoning action and the fact that a high enough water column can overcome athmospheric pressure. However the big IF is whether indeed a siphon action will take place and cause a vacuum at the top of the sealed container instead of break down the water continuity and cause a vacuum at the top of the tubes.
Siphons in vacuum work but combining that with a water column high enough to overcome atmospheric pressure is new to me. If it indeed creates a vacuum at the top of the sealed container then you can guess yourself what benefits that has.
33-34 ft is the magic number. at about there the "water continuity" you talk about is actually cavitation. water can no longer stay together. if less than that height, no worry. the weight on both sides (to the water level of the tank) of the tube will be equal. as long as the exit tube is lower than the top of the tank level, flow will happen provided the tank is vented (air can come in). if you plug this vent, flow will stop and leave a partial vacuum in the tank. depending on how far below the water level the exit tube ends will determine the amount of vacuum. as mentioned above, the height above the water level (up and down) in the tube will cancel each other.
at this point if you attach another tank, provided the new tube is full of water too, that is below the exit tube of the first tank, water will flow back to that (lowest) tank. this will cause the upper exit tube to empty. if the 2nd tank is higher than the end of the 1st exit tube, and vented, the water will flow out the 1st exit tube.
let me know if anything is not clear.
tom
QuoteP-Motion
elite_member
Hero Member
Posts: 1437
https://sites.google.com/site/besslermotion/
Re: Self Siphoning Water
« Reply #19 on: Today at 10:00:17 PM »QuoteQuote from: FreeEnergy on Today at 06:52:13 PMwhich way is the water flowing?
--edit--
never mind, i see it now!
when the water reservoir fills up to a certain water level the water gets flushed!
Hi Free,
The tall tube on the left would be for back filling it. I modified the drawing to have less waste in it.
Water would flow counter clockwise. One thing this would be testing is if vacuum allows something like this to follow hydraulic theory.
Not sure who in here is familiar with hydraulics. The basic idea is that the 5 pounds of water in the largr diameter tube would be on a piston that is connected to a piston the one gallon of water is on. This would focus the force of the 5 pounds of water and would push the 1 pound of water higher.
If vacuum allows for hydraulic theory, then the more water in the discharge tube should lift the lesser amount of water in the inlet tube.
according to the last pic, i would say this would not stay static. both pipes would drain. i believe the 5sq.in. still will only allow <1psi. since the pipe on the right is longer (and has 1psi), it will be the one to start the flow back. with this help and the large dia of the left pipe (no help with capillary), air will find its way in and the water in the left pipe will now drain. now you end with empty pipes and a tank of water going nowhere.
how close am i?
tom
have a look at this....
http://en.wikipedia.org/wiki/Heron%27s_fountain
from the page....
"Various versions of Heron's Fountain are used today in physics classes as a demonstration of principles of hydraulics and pneumatics."
i am starting to see my system not work.
the top water reservoir will empty out and the bottom tube will overflow with water :-(
i just have a gut feeling that the output siphon hose and the input capillary hose must be equal in diameter. it need to putout the same amount it puts in else it will overflow the system making it stop.
there must be a large water reservoir on top,...
this is just my opinion and no test(s) has been conducted yet.
someone may try it before me.
i know siphoning, tests must be made.
post results asap! :-)
keep it open source!
and the money should be either for the entire community as a means of improving the community as a whole, or this individual project alone once it works.
Too bad its not possible, pressure depends on head not area. but please give it a try. splice two hoses of differant diameters together and give
it a try.
fritznien
Quote from: fritznien on August 20, 2010, 10:39:05 PM
Too bad its not possible, pressure depends on head not area. but please give it a try. splice two hoses of differant diameters together and give
it a try.
fritznien
@everybody
will this help?
-edit-
did a small change to attached image.
maybe even like this...
?
isn't a source of vacuum an outside source of power?
you want something that will cause water to run up hill.
out side of a pump which takes power i think your out of luck.
but if you think the size of the hose will have an effect it should be cheap and quick to splice two differant size hoses together
and try it with a couple buckets of water.
or have i completely misunderstood?
fritznien
Quote from: P-Motion on August 21, 2010, 04:42:55 PM
Fritz,
The way I would be trying it at first, is to find out how much vacuum the static head for the inlet needs.
And then I would find out how much vacuum the discharge can create. If the discharge can create more vacuum, that would be the source.
If not, then it would not work. A vacuum gauge with pump costs $30. With something like this, no guessing would be needed. Different sizes of tubing or pipe could be tested.
By having a reservoir on top, it nicely seperates the inlet and discharge. This would create a nice visual effect.
i still like my idea, some tape and a couple of pieces of scrap poly pipe would show proof of concept in 20 minutes of work.
as for the vacuum you need thats going to depend on the hieght. but why not do what many people do and just fill the unit with water and cover the ends of the hose till its all in position.
that works well for a single siphon.
but what do i know.
fritznien
**** Deleted ****
very good!
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
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
Quote from: P-Motion on August 20, 2010, 08:34:14 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
See the Open Source Vs. Patenting topic over here: http://www.overunity.com/index.php?topic=1821.0 (http://www.overunity.com/index.php?topic=1821.0)
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.
QuoteThis 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
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?
QuoteAs 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
Quote from: P-Motion on August 24, 2010, 07:28:01 AM
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.
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.
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
Quote from: Low-Q on August 27, 2010, 05:28:16 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
Anyone?
whats to say he is talking about a pump.
fritznien
Quote from: fritznien on August 28, 2010, 10:27:59 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