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Imagine we have a vertically orientated pressurised liquid filled closed pipe.

The pipe is shaped like a spring, such that the distance between the two ends can be adjusted vertically, whilst the pipe remains liquid filled.

If the distance between the ends of the pipe is adjusted vertically;

Will

a) The pressure change only be at the lower end of the pipe, varying according to $\rho gh$, with the pressure at the upper end of the pipe remaining unchanged?

Or will

b) The pressure vary at both ends of the pipe?

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Problems where a (non-compressible) liquids fill a specific volume are ill-defined. That is, an infinitesimal volume change may affect the pressure dramatically.

There's a classic problem when there's a tank filled with liquid completely, except there's a little air bubble at the bottom of the tank. Now, what happens when this bubble eventually detaches from the bottom and floats to the upper bound of the tank? The answer is that the pressure that was earlier at the bottom will now be at the top, and the pressure at the bottom will increase according to Archimedes' low.

This is an example of an ill-defined problem. What'd happen in real life is that upon the pressure increase the tank would expand a little, the volume of the bubble will grow, and the pressure of the liquid will barely change.

So, regarding your question: it's clearly ill-defined. It's impossible to determine the liquid pressure from the beginning, it's unclear if the overall volume of the pipe changes upon its deformation. The only thing that is known is the pressure difference at top/bottom, which is according to Archimedes' low.

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  • $\begingroup$ Hi, yes sorry for being a bit unclear. The hypothetical situation assumes that there is zero change in volume as the distance between the ends of the pipe are changed. $\endgroup$ Oct 7 '15 at 6:22
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You're changing the volume of the pipe, right? by changing its length but not its diameter? The pressure will vary at both ends of the pipe, because the only difference between the two pressures is due to the height of the fluid column.

Is the fluid incompressible, like water or hydraulic fluid is usually assumed to be? or is the fluid compressible, like air? If it's considered incompressible, what that means is it's very very stiff, so it takes a large pressure change to produce a small volume change.

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  • $\begingroup$ Thanks mike and valdo for your responses. In the hypothetical situation I am asking about, we imagine the volume of the 100% liquid filled pipe remains totally unchanged as the vertical distance between the pipe ends is varied (the spring shape of the pipe allows for this). I am trying to understand pressure changes due to elevation changes only $\endgroup$ Oct 5 '15 at 20:27
  • $\begingroup$ @JamesEadie: if the volume remains unchanged, that means as you stretch the pipe its diameter must get smaller (you said - no volume change), and that means the only pressure change will be at the bottom, because the height of the fluid column above it has increased. $\endgroup$ Oct 6 '15 at 0:13
  • $\begingroup$ Hi Mike, Yes in the hypothetical example we assume there is zero change in volume. I like you thought that only the pressure at the bottom of the tube will change. However, I have a colleague who says that pressure at top and bottom of pipe/tube will change (the difference between the values at top and bottom is still according to rho x g x h, its just that pressure change due to height change is split between top and bottom, with bottom end value rising and top end value dropping) $\endgroup$ Oct 6 '15 at 8:45
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The basic assumption about how the length of your tube in changing would be that each small section changes by the same amount $\delta y$ resulting in a total change $\Delta y$. With that reasonable assumption, the pressure in each small section will change $\delta p = \rho g \delta y$, and this will be uniformly distributed throughout the height. The total pressure difference between the top and the bottom will be $\Delta p = \rho g \Delta h$.

If the fluid is incompressible and the walls are perfectly rigid, you won't be able to move the coil. If the tube walls can flex in or out, the overall pressure may change, but it won't change at a single elevation. Any changes will be distributed based on how the length change is distributed.

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  • $\begingroup$ Thanks Bill. Yes the fluid is incmpressible and the pipe is flexible without volume change. I get that the pressure difference between the two ends of the pipe will allways be in accordance with rho x g x h. My question is will the additional pressure due to change in distance between the two ends be added on to the bottom end of the pipe, or can the pressure at the upper end of the pipe also change? $\endgroup$ Oct 5 '15 at 20:43
  • $\begingroup$ In my answer I said that the change will be uniformly distributed. $\endgroup$
    – Bill N
    Oct 5 '15 at 20:45
  • $\begingroup$ OK to check I understand; if we assume 10m head of water has a pressure of 1 bar. And, we assume the original pipe configuration has 200 bar at upper end, with 100m vertical separation between ends (hence 210 at bar at lower end). If we increase vertical separation by 30m, then bottom of pipe now reads 211.5 bar and upper end reads 198.5 bar, correct? thanks $\endgroup$ Oct 5 '15 at 20:57
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Imagine you "unwrap" your coiled tube, so it looks like a slanted tube. Every bit of the tube is the same height as before, but it is no longer going around in a circle.

Now "stretching" the coil is equivalent to tilting the tube more vertically.

It should be easy to see that this will change the pressure difference between the top and the bottom.

Whether this changes the pressure at the top, the bottom, or both, depends on whether there is any boundary condition that forces the absolute pressure to be constant at either of these points. If the top is open, the pressure there will be constant. If the tube is completely closed, then I believe the answer will depend on the Poisson ratio of the tube material: it is actually possible for the volume of the tube to become (very slightly) smaller or larger due to the bending and stretching operation.

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  • $\begingroup$ Hi Floris, The tube is completely closed and nothing is forcing pressures to remain constant ate either ends of the tube, and in the hypothetical example no change in tube volume. I get that moving the ends will change the pressure between top and bottom, and it seems that the change in pressure is just an increase at the bottom of the tube, with the top value remaining constant. Although it is this last part that I am not sure of $\endgroup$ Oct 6 '15 at 8:47
  • $\begingroup$ If the tube does not change volume, the change in pressure in the bottom will compress the liquid (a tiny bit) causing it to occupy a smaller volume and lowering the pressure at the top. But that may be overthinking the problem. I think it is worth a conversation with your instructor (I imagine this is a homework problem...?) $\endgroup$
    – Floris
    Oct 6 '15 at 10:30
  • $\begingroup$ Not a homework problem, but a conversation at work. $\endgroup$ Oct 6 '15 at 10:53
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I clarified this further with my colleague (PHD in physics and works with fluid dynamics).

Wierdly (to me anyway) apparantly the pressure will change at the top and bottom of the pipe.

The difference between top and bottom varies in accordance with rho x g x h, however the average pressure (the pressure at the vertical midpoint) always remains the same in a closed system.

He made an OLGA model to demonstrate this.

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