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closed tank ... straw effect

  • Thread starter Thread starter mir
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mir

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Hi.
I imagine you all know the game with the closed straw that does not bring out water if I stop one of the ends ... the question that as a child appeared to me with the characteristics of magic now, which are grown, reaches the level of persecution :) ... I explain:

I have a tank at 4 meters height
phase 1: pomp in water along the valve pipe
phase 2: I close the valve and "I'd like" that the tanks empty but I fear the straw effect and then I would like to put a depression valve on the tank but... what depression do I create in the tank?

I ask the needle for a little help.
Thank you.
♪
 

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I would say that if you can fill the tank (and then the air comes out somewhere), you will also be able to empty it (the air will come back from where it came out).
the "cannuccia effect" as you call it is simply due to the fact that the ancients perhaps did not have all the twists with the concept of "horror vacui". it is not healthy to try to put the tank in depression, just one vent up.

If you explain more in detail what you want to do, maybe we better understand why such a strange scheme. in any case the pressure in the tank will be equal to the atmospheric one less the column of water (four meters plus the height in the tank). If the pipe was 10,33 meters, the water in the tank would boil. . .
 
as well as schematized, assuming a perfectly stagnated tank, there is no straw effect!
In fact, when you blow the water, the pressure in the tank will rise to levels so high to shoot you out the water even without calling in cause gravity....

For the rest... I agree with lightning
 
ok I could put the vent but I fear that the water comes out from the vent when I pompom inside (as the pump covers various utilities there is the risk that I enter to 3 or 6 bar and therefore my tank becomes a fountain)

from what you tell me then the pressures are:

1 atmospheric bar - 0.4 hydraulic bar = 0.6> absolute ... is it -0.4 bar?

if I mount a depression valve at 0.02 bar opens quickly and allows the outflow.

What do you say?

♪

ps: actually at 1 absolute water would boil at 45°c ... http://it.wikipedia.org/wiki/ebollizione
 

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MBT said:
as well as schematized, assuming a perfectly stagnated tank, there is no straw effect!
In fact, when you blow the water, the pressure in the tank will rise to levels so high to shoot you out the water even without calling in cause gravity....

For the rest... I agree with lightning
Click to expand...
Yes but the problem I can imagine just when I don't pompo... unfortunately I have low ceiling and can't put a high vent :
 
Enigma said:
Excuse me the question... must it be closed?

enigma
Click to expand...
no but I can't put a very trivial vent (see previous posts) and I have to put a depression valve but I wanted to be sure to correctly calculate the depression...
 
my post before was more done than serious....
He wanted to say that if the tank was stagnated you would put it in pressure when you introduced us water and solve the problem of depression. . .

speaking more practically.. .
the "problem" lies in the speed in which you make water flow and flow.
I'll explain.
If you work slowly, a ø2 dandruff may be sufficient at the top.
from there the air enters and exits, balancing the pressure
the water from a 2-hole comes out in minimum quantities, you will see it and you have time to stop the pump without having to swim towards the switch.
If you need high flow rates, the "hole" must be 1/2". then either put some kind of valve or you need a sub mask to go off the power pump.

the choice becomes according to the situation at the contour
Could you use a vented cap like this?
http://www.elesa.com/scheda_en_1_7601_1_4.aspxor make one you, inside the tub, with a hose and a floating ball
 
mbt I prefer the valve otherwise I set a web cam and remotely look if it comes out ... and now as the valve size ... is correct the calculation I made above?
 
No.
you have to consider the volumes.
I mean, how much water do you want me to go out before the valve intervenes.
As I understand that this tank is an accumulation and you want to empty it by gravity, maybe you could use a solution of this type:
 

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the tank holds 20 lt and I care that you download them in 2/3 minutes (sui 400 lt/h).

from the scheme I see a pressure gauge with a left pump + a depression valve at the top + a valve at the exit?
 
usually, I use what "I opened on the desktop"
So... pump with motor (not gauge) to indicate that from I will pump water in the tank
on the other side, willow (so drain when and how much I want)
above, valve made in the house.
in synthesis, floating spherical ball, housed inside a smooth tube of diameter 1-2 mm higher
closed under perforated plate
above the sphere, conical seat with passing hole
water floats the ball
the ball, floating, reaches the hole
the ball, supported by the water, closes the hole and prevents the water from running out.
When I open the water, the water comes out anyway
the water, dropping level, drops the ball that opens the hole that balances the pressure
 
mir said:
the tank holds 20 lt and I care that you download them in 2/3 minutes (sui 400 lt/h).

from the scheme I see a pressure gauge with a left pump + a depression valve at the top + a valve at the exit?
Click to expand...
a 20 litre tank is a water tank!!!! :eek:
I thought it was at least 1 metrecubo stuff.
 
mir said:
the tank holds 20 lt and I care that you download them in 2/3 minutes (sui 400 lt/h).
Click to expand...
forgives irony (but not too much).. but in your case it would soak a very trivial "basket of a basket".
more "industrially" there are on the market of the tanks, galvanized or stainless, with floating filling, open sky with lid. you put what you want to unload; simple and effective.
 
sampom said:
forgives irony (but not too much).. but in your case it would soak a very trivial "basket of a basket".
more "industrially" there are on the market of the tanks, galvanized or stainless, with floating filling, open sky with lid. you put what you want to unload; simple and effective.
Click to expand...
I know that would suffice that ... but also banally a valvolina di ritegno opening at 0.025 bar ... but I still do not know how to calculate the depression that is created!
 
mir said:
...but I still don't know how to calculate the depression that you create!
Click to expand...
How come? What doesn't convince you of what I wrote?
Fulvio Romano said:
[...]If you explain more in detail what you want to do, maybe we better understand why such a strange scheme. in any case the pressure in the tank will be equal to the atmospheric one less the column of water (four meters plus the height in the tank). If the pipe was 10,33 meters, the water in the tank would boil. . .
Click to expand...
 
then it is correct what I said:

1 atmospheric bar - 0.4 hydraulic bar = 0.6> absolute ... is it -0.4 bar inside the tank?

if I put a depression valve at 0.02 bar opens quickly and allows the outflow?
 

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