Continuously Flowing Water Theory

[CompuTutor]

If you watch the Video below, you will see Continuously Flowing Water WITHOUT a VACUUM AND Vacuum Pump.
It could be Scaled Up so the falling water could drive a Water Wheel that turns a Generator.

FREE ELECTRICITY forever.  Why monkey around with a vacuum???

http://www.youtube.com/watch?v=287qd4uI7-E&feature=channel&list=UL

You guys do realize this is just a nice parlor trick,
involving electrostatic liquid attraction as a pump,
right ?

Look at the conveniently placed two upright stands,
that spacious base, and think it through this time...

It wouldn't matter if the left side was the size of an ocean,
and the right side just a small garden hose in comparison,
they would (only) equalize levels, that's all...



EDIT:
Heheh, after looking at that for a full three seconds more,
I spotted an easier way, tap the liquid at the left stand's bottom support,
use a pump underneath, and a venturi in that right stand's support.

You did you all notice how no liquid even came out that tube
at the bottom of the flask right away like it would normally,
the flask is 1/3 full before anything starts to flow into it ?

That's because the suction of the pump is taking it instead.

That rules out electrostatic pump,
and rules in standard liquid pump,
with a venturi in the right support.

Bet there is several other ways to do this too,
but this isn't worth more than thirty seconds worth of thought...



EDIT 2:
I have now read the rest of these two pages,
(I have my posts-per-page set to maximum...),
enough people saw the vid's inconsistencies.



I'm was sad to see mechanical advantage added at first,
I'm trying to warm up to this combination though.

An initial pressure differential to utilize,
combined with controlled cyclic valving added
is where I was hoping this would go instead.

I also didn't see any consideration to a time interval.

For flow to double, the pressure differential must quadruple,
for both the pressure head heights to acheive double the flow.

Like 'Micro said, some form of flow control is needed,
then the resulting equalization unbalanced again,
rinse and repeat...

At least there is a constant to work against:
Each 2.31 feet of head height differential = 1 Pound (per) Square Inch (PSI) increase.

If only there were a way (bouyancy?) to juxtaposition the relational height
of each of the two chambers, and hence their current water head heights
in relation to each other with less work input somehow ?

After all, a perfect vacuum will hold a 34 foot high head of water,
that must be useful in application to developing a working concept...

The vacuum doesn't have to be "Produced" either,
a submersed unit, allowed to vacate of any air first,
need only now be raised to begin the vacuum process.

Then, inevitably, the suspended gasses in the liquid water
would release and provide the small gap to start the process.

I Like the logic problem presented though,
going to mull this over for a few days now.

Thanks Johnny, be well.

(Others, please don't quote this entire lengthy post, trim it please, TY)

[Lakes]

I was wondering when someone would spot the "thick" base that (youtube) demo is on...

[CompuTutor]

There is another thing that makes me wonder about standards,
why are basically all incompressible liquids measured in PSI ?

We use PSI as a square of the pressure with compressable air as example,
it isn't just per-square-inch, it is a square of the pressure per square inch,
a quadrupling of occupation for every doubling of the pressure.

For those not getting that, think of a printer's DPI (Dots Per Inch)
10x10 = 100-DPI
20x20 = 400-DPI
lousy example, but good enough to visualize compressables like air,
increasing from 10 units of pressure to 20 units, quadruples air stored.

But as water is basically nearly incompressible,
why isn't it just measured as "Pounds" of pressure,
like inches of mercury or hg for a vacuum as example ?

I bring this up because in order to double the vacuum available,
quadruple the amount of gasseous substance must be removed.



Also, for some reason, double/triple/quadruple steam engines come to mind,
probably from the 2:1 ratio your working against in your ideas Johnny.

Perhaps the opposite could be employed to harvest the most effective force,
each chamber being 2:1 larger to reacted to half the pressure with equal force ?

(Note 2:1 ratio from chamber to chamber in pic, then reverse that thought for vacuum)

NOTE:
I just attached an animated graphic interchange file (Ani-GIF),
and the forum resize script seemed to have de-animated it 
(to a single-framed picture instead...)

Download to see animation instead:
http://www.overunity.com/downloads/sa/downfile/id/500/
(Clicking the pic link gets the same single frame pic sadly...)

[excessAlex]

What is your native language if you don't mind my asking ?..

....something working with a liter or 2 would be enough to impress most anyone

I am Italian  Nice to meet you

I tried to use some little reservoirs ( e.g. 1,5 liter ), and the result are not so good, because the pipe for the rising fluid must be very little and the friction will be very strong. I I need a tiny tube that must be, however, very, VERY smooth on the inside. For this I thought to enlarge the size (and weight) of the whole device so I could indirectly "shrink" the imperfections of the pipe internal  .. At least that I thought in my ignorance 

.. For the lever:
I had thought of a lever, but then this lever interferes with the lifting of the low-intermediate vessel. There is a margin in order to use a little leverage, but not much.
I thought for example to add weight to the container upper-intermediate, for example: if the containers were to hold 100 kg of liquid, the weight to be added may be up to 80 kg, because the container medium-low, when the liquid is transferred to its inside then it would weigh 100 kg against 80 kg of weight added to the container upper-intermediate .. This could be turned into a lever, rather than add weight so you could change the arm that regulates the movement of intermediate containers ..

Alex

[johnny874]

I am Italian  Nice to meet you

I tried to use some little reservoirs ( e.g. 1,5 liter ), and the result are not so good, because the pipe for the rising fluid must be very little and the friction will be very strong. I I need a tiny tube that must be, however, very, VERY smooth on the inside. For this I thought to enlarge the size (and weight) of the whole device so I could indirectly "shrink" the imperfections of the pipe internal  .. At least that I thought in my ignorance 

.. For the lever:
I had thought of a lever, but then this lever interferes with the lifting of the low-intermediate vessel. There is a margin in order to use a little leverage, but not much.
I thought for example to add weight to the container upper-intermediate, for example: if the containers were to hold 100 kg of liquid, the weight to be added may be up to 80 kg, because the container medium-low, when the liquid is transferred to its inside then it would weigh 100 kg against 80 kg of weight added to the container upper-intermediate .. This could be turned into a lever, rather than add weight so you could change the arm that regulates the movement of intermediate containers ..

Alex

   Hi Alex,
With leverage, it increases the force acting on the pump, very helpful.
With what you have shown and what some might miss is that by increasing the diameter of the
pump, it's height can be less. If adhesion or cohesion is a problem, then some light oil can be
added to the water to coat the inside of the tubes.
With the riser pipe, it might help to calculate the volume the pump holds and then this would
help to determine what size riser pipe might work best. PVC is pretty smooth on the inside
and is used agreat deal in plumbing.
To caclulate volume of a static head or riser tube, the formula is PiR^2H or 3.142*radius^2*height.
^2 represents squared or multiplied by itself and * is times. Just so everyone knows 
By the way Alex, the fulcrum for leveraging the pump could be suspended from the bottom of the
top reservoir. Also, if volumes are known, then the total heaight of your design could be lower than what
you might think. If so, then the diameter of the riser pipe might be able to have a larger radius.

                                                                                                                          Jim