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Main question: If you have a column of fluid with a valve as shown on the drawing, what happens to the pressure at the point p when moving from state A to state B and vice-versa?

I'd think it gets suddenly decreased because now it's at a much smaller depth, but I have trouble dealing with the idea of such a sudden change.

Bonus track: What could one say about the force required to open or close that valve? On the 2) section of my silly drawing: what should K be? How close or far away from 1 should it be?

schema

Some thoughts about this problem:

  • When closing the valve, the pressure is even aboe and below the valve level, so I'd think closing the valve doesn;t require lots of force. At least, that force shouldn't be depending on the height of the container.
  • If the pressure does change from one state to the other, then opening the valve could be tricky. The big pressure difference between sides of the valve could cause a problem there. Maybe the required force will be very high (height depending) or maybe the thing would explode (like suddenly opening a door on a flying plane).
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2 Answers 2

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Tricky question (well a little bit tricky).

But the correct answer is - no change in pressure. Assuming the valve is closed very slowly so that it does not act to displace water as it closes, and that the valve creates a perfect seal as it is closed, the water in the container below is essentially trapped at the original pressure, $P=\rho g(H+h)$. When the valve is closed, the lower part of the container is no longer affected by the column of water above it, but it does retain the pressure from the original pressurization.

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    $\begingroup$ It makes sense, this way there's no sudden pressure change which could lead to a explosion or something like that. It would be strange to trigger an explosion (or setting the energy on place for it) from the simple act of closing a valve. This means that K is aprox. 1 since on both scenarios the pressure is aprix. even at top and bottom of the "valve area". Thanks. $\endgroup$
    – daniloquio
    Dec 4, 2015 at 21:11
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    $\begingroup$ The thing that matters is the pressure difference across the door (valve). As long as you keep that near zero then very little pressure force. But if you move the door too quickly you might be able to create pressure differences due to drag forces. And this might lead to a slight reduction in pressure below the door from what you started with. The trick is to keep the fluid velocities low as the door is being closed. $\endgroup$
    – docscience
    Dec 4, 2015 at 21:16
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1) If you have fluid in container the pressure does change and the amount of change is teh same as the formula that you have in (1).

2) The amount of the force in order to close the valve (if we assume the valve as a divider that you can abruptly place it and not think of a hinged lid or sliding a divider into its position and...) is:

$F=P\times A$

where $P=\rho g H$ and A is the area of the top part of container (the narrower part). In my opinion you do not need any force to open the valve. As soon as you release its stopper the fluid does the job itself with a force that is equal to the closing force. So K should be equal to one.

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  • $\begingroup$ Could you elaborate on why 1) is correct? I really find it hard to believe because all that pressure cannot disappear just like that. Also I have trouble following the last part because is pressure changes, opening the valve wouldn't be that simple because there's a pressure difference there. $\endgroup$
    – daniloquio
    Dec 4, 2015 at 21:15
  • $\begingroup$ Because pressure is nothing other than a form of force. When you shut the valve it bears the amount of force (F=$P*A$) from the top column and prevents it from being sensed by the fluid underneath. Then it is as if you remove the top part. The amount of force you need to shut or open the valve is defined by the pressure at that point (in fluids you speak of pressure easier than force) which no matter if you want to turn of the valve or turn it on is the same and equal to $\rho g H$ $\endgroup$
    – Amin
    Dec 5, 2015 at 7:23

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