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Here is something I see :

enter image description here

Let's say the ideal fluid(water here) of density $\rho$ is drawn from a source by a motor and thrown upwards with a velocity $v$. Now we take the power of motor be constant and fluid to flow in streamlines (same velocity all over because of same cross section area) ,then applying Bernoulli's Equation for two points shown we get the $$P_{down}-P_{up}=\rho g h$$

$P_{down}=P_{atm}$ because the down point is open to air. So, we get $$P_{up}=P_{atm}-\rho g h$$

which will go negative if $h>10m$. How is it possible?

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This: youtube.com/watch?v=BickMFHAZR0 suggests that with no air bubbles cavitation will not occur and negative pressure will be observed. this is how trees pump the water to the top. –  user31353 Oct 19 '13 at 20:12

3 Answers 3

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which will go negative if $h>10m$. How is it possible?

It is not possible that $P_{up}$ will go negative. It will experience cavitation. This is one of those cases where you have to realize that the equation you're using is a special case of a larger theory, step back, and employ the larger theory.

If you step back from the concept of water as an absolute liquid, then it's clear that it has a state diagram. Going from some liquid place on the state diagram to $P=0$ will cross a phase transition line. Probably just once, and probably the liquid to vapor line. That means the water boils.

As you hit this threshold of $10 m$ between atmosphere and the lowest pressure point in the system, that point will turn to gas. There will possibly be some significant change in temperature, but we can imagine it happens slowly so everything is held at room temperature. So draw a line on the PT diagram of water of constant T, decreasing pressure.

I have to make some adjustments to your diagram now. Firstly, you imagined an open tube pointing down. I'll have to revise this to a tube that snakes around so that it has a defined level. Take a pipe filled with water and turn the open end down - the water is likely to fall out. We can often ignore this in small (common aquarium) pipes because of surface tension effects.

In addition to that, let's imagine the cavitated area that is now filled with water vapor. I'll, of course, assume your pipe is strong enough. Not to worry, plenty of pipes are.

cavitated area

There's some detail that I didn't include. But let's start making observations:

  • The pump's exit is greater than atmosphere pressure, because the head above it is $>10m$
  • The white areas in the pipe are filled with water vapor, in fact, saturated water vapor at room temperature
  • The pressure of the vapor region will be the saturation pressure at room temperature. This is not zero, but it is very low.

As the pump pushes more liquid through the system, that liquid will overflow through the vapor region and fall down, joining the water in the outlet pipe.

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So, all water in pipe above $10m$ height will be gas?? Does not seem correct. The water tanks on top of our homes, we get them filled with help of motor only and it get's in water, not vapors. –  ABC May 24 '13 at 2:42
@007 The motors can pump any height you like, they just need to be at the low elevation point. It's standard engineering thinking to say "if you pump up a hill, put the pump at the bottom". For a home (this is somewhat of a guess) I imagine that the pipes should be above atmosphere everywhere. –  Alan Rominger May 24 '13 at 12:23

The pump can apply only a fixed pressure/force from downwards . From the open end the atmosphere will apply a fixed force/pressure .

It is only the height of the column which is variable here . If you need height $>10$m , you just need a bigger pump . If you can't get that , the water will rise upto a fixed height say $8$m,after which there will be hydrostatic condition achieved and fluid won't rise any further , pressure can't be negative under any circumstances as pressure isn't the variable here , height is . And height has to determined for a fixed pump pressure .

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This realationship:

$$P_{down}-P_{up}=\rho g h$$

will only hold for a standing water column. Definitely not for a free-falling one.

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I got this by applying Bernoulli Eq. , $v$ at both points is same. –  ABC May 24 '13 at 2:35

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