Propagation speed of movements of a fluid in a pipe

[coquelicot]

TL;DR Summary

Water in a long pipe is suddenly strongly pushed at one extremity during some time. After how many time will it begin to move at the other extremity?

Assume that we have a 1.5 km x 100 cm^2 long straight pipe, totally inelastic and full of water. From time t = 0, a pressure of 300 000 Pa is continuously applied to the water with a piston at one extremity. This correspond to a force of 30 000 N on the pipe cross section in the direction of the other extremity of the pipe.
Notice that the total mass of the water in the pipe is 15 000 kg.
Also, the speed of the sound in water is approximately 1.5 km/s.

Neglecting the friction of the water on the walls of the pipe, basic Newton laws say that in one second, the piston should move at least by 1 m in the direction of the other extremity.
The question is: at what time t will the water begin to move at the other extremity?

Note: if the moving signal were propagating at sound's speed in water, then the water would begin to move at time t = 1 s at the other extremity. This means that during this time, 1 m of water would have been compressed inside the pipe. This sounds to me like an absurdity, because the applied pressure is insufficient to do so.

 

[jrmichler]

This sounds to me like an absurdity, because the applied pressure is insufficient to do so.

Have you done the calculation for compressibility of water? You made the simplifying assumptions of a rigid pipe, zero friction, and constant pressure suddenly applied at one end. Now make two more simplifying assumptions: the pressure is slowly increased from zero to 300,000 Pa, and the far end of the pipe is blocked. Use the actual bulk modulus of water to calculate just how much the piston will move.

In this case, "pressure is slowly increased" means that the pressure is increased slowly enough that there will be no dynamic effects.

 

[Dale]

The question is: at what time t will the water begin to move at the other extremity?

1 s

This sounds to me like an absurdity, because the applied pressure is insufficient to do so.

A 1 m compression of a 1.5 km object is just a factor of 0.00067. I am not sure why you think 30 kN is insufficient for a 0.00067 strain. Can you provide your supporting calculations

 

[Baluncore]

This sounds to me like an absurdity, because the applied pressure is insufficient to do so.

What kinetic energy is stored in the moving water at the instant it is stopped ?
Where does that energy go, and how does it get there ?

 

[coquelicot]

What kinetic energy is stored in the moving water at the instant it is stopped ?
Where does that energy go, and how does it get there ?

No kinetic energy and the water was not stopped. The water was at rest before time = 0. The water moves because of the piston at one extremity.

 

[coquelicot]

Have you done the calculation for compressibility of water? You made the simplifying assumptions of a rigid pipe, zero friction, and constant pressure suddenly applied at one end. Now make two more simplifying assumptions: the pressure is slowly increased from zero to 300,000 Pa, and the far end of the pipe is blocked. Use the actual bulk modulus of water to calculate just how much the piston will move.

In this case, "pressure is slowly increased" means that the pressure is increased slowly enough that there will be no dynamic effects.

OK. Here is how I performed my calculations (I may be wrong though).

The total volume in the pipe is $15 m^3$. The pression is 300 kPa by hypothesis. The bulk modulus of the water is K = 2.2 GPa. We have $$K = -V dP/dV.$$
Since V is sensibly constant during the process, I wrote
$$K = - V \Delta P / \Delta V = -15 \times 300 000 / \Delta V.$$
Hence $\Delta V = -0.002045 m^3$. Since the cross section of the pipe is $100\ cm^2 = {1\over 100}\ m^2$, $\Delta V$ corresponds to a length of pipe equal to $100 \times |\Delta V| = 20.45\ cm$.

EDIT: I had a stupid calculation mistake inside the question itself: the force corresponding to 300000 Pa is not 30 kN but 3 kN, because the cross section of the pipe is ${1\over 100} m^2$. This reduces the length of the piston movement by a factor of 10: not 1 m, but 10 cm. Now everything makes senses. Sorry for the time wasted by you and the other the readers.

 

[coquelicot]

1 s

A 1 m compression of a 1.5 km object is just a factor of 0.00067. I am not sure why you think 30 kN is insufficient for a 0.00067 strain. Can you provide your supporting calculations

Yes. See my answer to jrmichler. thx.

 

[jrmichler]

Now everything makes senses. Sorry for the time wasted by you and the other the readers.

You asked for help. That help enabled you to take a different look at the problem and solve it. Nobody's time was wasted. Not yours, not ours.