Re: Free energy

In sci.physics, Gieniu
<warendag@xxxxxxxxx>
wrote
on Sun, 17 Jun 2007 23:33:30 GMT
<exjdi.30608$nx3.23165@edtnps89>:

"Gieniu" <warendag@xxxxxxxxx> wrote in message
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"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
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In sci.physics, Gieniu
<warendag@xxxxxxxxx>
wrote
on Mon, 04 Jun 2007 20:18:38 GMT
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"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
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In sci.physics, Gieniu
<warendag@xxxxxxxxx>
wrote
on Sun, 03 Jun 2007 06:08:45 GMT
<NVs8i.67391$V75.42901@edtnps89>:
Hi
The pumping it is moving of liquid from the lower level ho, to .
higher level h1.
Because this action is in the gravitational field,
therefore is prformed the work W=mg( h1 - h0),
and this work the motopump performs.
The straight tube is plunged deep into ideal liquid,
This tube is open on the both its ends.
On the top of tube is instaled the motopump
and the lower end of tube is open.
By action of the hydrostatic preasure, the liquid
flows into
the tube and into the motopump.
Maximal velocity v of pumped liquid is
v = sqrt ( gH)
Velociti of liquid out of pump is also same
but here there are two different action,
1/ keeping velocity of liquid as constant,
2/ hydrostatic preasure.which creates of
velocity of water equal v=sqrt(gh).
These two action are self-supporting

In order to receive energy E = m vv/2 =mgH/2
WE must give energy E1=mg(h1-h0);
it does the motopump
This work is performed for EVER velocity
of pumping ideal liquid ,on the condition that
this velocity is constant, it is not variable.
.
Sincerely E.W.


The bottom of the tube does not is empty ,when the water
flows into it.

--
#191, ewill3@xxxxxxxxxxxxx
Does anyone else remember the 1802?

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*********************************************
Hi
This tube is not empty ,it is full of the
water(ideal liquid) when motopump is working.
Sincerely E.W.


Then I have no idea where your motopump is. If it's at
the top, then it can move fluid around but that's about it.

If it's at the bottom, it's under tremendous but balanced
pressure from both sides of the tube bottom and can move
fluid around but that's about it.

--
#191, ewill3@xxxxxxxxxxxxx
Linux sucks efficiently, but Windows just blows around
a lot of hot air and vapor.

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Posted via a free Usenet account from http://www.teranews.com




*****************************************
Hi
The motopump is intalled on top of the tube.
The lower end of the tube is open
The liquid fiows into the tube ,across the lower end fows out
of it into the motopump with the speed of v = sqrt (gH)

Um...where exactly did you get that notion? The flow of
the liquid is going to depend on the power supplied to
the pump, minus any efficiency issues (e.g., viscosity
and friction between the tube sides and the liquid).

Were your relation true you'd be able to observe some
interesting things using little more than a soda straw,
about 10 cm (0.1 m) in length. With your equation,
v = sqrt(9.805 * 0.10) = about 1 m/s, which would result in
a continuous geyser about 5 cm high. Stick the soda straw
into a sufficiently tall water glass and see what happens.

I predict nothing will happen. :-)


Sincerely E.W.




--
#191, ewill3@xxxxxxxxxxxxx
Conventional memory has to be one of the most UNconventional
architectures I've seen in a computer system.

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Posted via a free Usenet account from http://www.teranews.com

.

Relevant Pages

  • Re: Free energy
    ... The pumping it is moving of liquid from the level ho, ... This tube is open on the both its ends. ... On the top of tube is instaled the motopump ... Maximal velocity v of pumped liquid is ...
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  • Re: Free energy
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  • Re: Free energy
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  • Re: Free energy
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