-1
$\begingroup$

I was curious to know does water come out of a certain depth when there is a difference in pressure, and also if it does come out of a certain depth, at what velocity does it travel at?

enter image description here

So, basically Valve A is opened when the tube has reached a depth of 100m under the sea and the tube does not gets crushed.

My questions are as follows:

  1. If valve A is opened when the tube reaches a depth of 100m then will the water travel in the upward direction as marked by the blue arrows?
  2. If the water does travel in the direction of the blue arrows, at what velocity does it travel?

It would be of great help if anyone can share me some links related to this topic.

Thanks in advance!

$\endgroup$

3 Answers 3

1
$\begingroup$

It depends on how wide is the pipe is and how much water it displaces will dictates how fast the water would equalize. Due to stored energy from the pipe being held under water then being released into kinetic energy it may splash out the top but not keep flowing out the top regardless of the pressure at the base of the pipe. The water levels will be the same inside and outside the pipe.

$\endgroup$
0
1
$\begingroup$

Before valve-A is opened, the pressure in the tube is atm. Pressure since the top of the tube is opened. When the valve-A is opened, assumed instantaneously, the water will rush into the tube at:

$$V = \sqrt{2gh}$$ where: $g$ = gravitation const (32.2 ft per sec per sec.), $h = 100\ \mathrm m$ = Height from bottom of tube to tank water level.

Unless the top entrance of the tube bends down below the tank water level, the water will only rise to the level in the tank.

$\endgroup$
1
  • 1
    $\begingroup$ actually, the water will probably not come out even if "the top entrance of the tube bends down below the tank water level" ... if it had once passed, it would continue to flow out, but it would in most scenarios not have enough inertia to rise over the water level far enough to reach the top point of the tube. $\endgroup$
    – Ilja
    Commented Apr 23, 2016 at 19:21
1
$\begingroup$

The real answer is quite complex; I think we should break it into a couple of different pieces.

First - the static case. If you submerge an open pipe into water, the pressure inside and outside will be the same at a given height, and the water level inside the tube will settle at the same height as outside. If you add the effect of surface tension, it is possible (depending on the material and dimensions of the pipe) for the water level inside the pipe to be either slightly higher, or lower, than the water level outside.

Second - the dynamic case, ignoring friction. When the valve is opened, the water under high pressure at the bottom will rush into the pipe, and the column of water will have work done on it by the pressure until it is level with the water outside. However, it also has to do work against gravity; we can calculate the velocity of the water by looking at conservation of energy. Assume length of pipe $\ell$, area $A$, filled to a height $h(t)$ at time $t$ with liquid of density $\rho$, with acceleration of gravity $g$. The pressure at the bottom of the pipe $\Delta P = \rho g \ell$, so the total work done to push water all the way up is

$$\begin{align}W &= \int \Delta P A dh \\ &= \rho g A \int\ell dh \\ &= \rho g A \ell^2\end{align}$$

Meanwhile, the liquid was raised to an average height of $\ell/2$, so the work done against gravity was $m g \ell/2 = \rho A \ell g \ell/2 = \frac12 \rho g \ell^2$.

The difference is kinetic energy of the column, meaning that the water is indeed moving when it reaches the top of the water surface:

$$KE = \rho g h A (\ell^2 - \frac12 \ell^2) = \frac12 \rho g A \ell^2$$

Since the column of water has mass $m = \rho A \ell$, we find the velocity $v=\sqrt{g\ell}$.

The height that this water can reach is determined from simple equations of motion. First, let's solve this if the pipe ends at the surface. In that case, water moving at velocity $v$ will take $t=\frac{v}{g}$ to reach a maximum, at which point it has travelled a distance $y = \frac12 g t^2$.

Substituting:

$$\begin{align}y &= \frac12 g t^2\\ &= \frac12 g \left(\frac{v}{g}\right)^2\\ &= \frac12 g \frac{g\ell}{g^2}\\ &=\frac12 \ell \end{align}$$

A surprisingly simple result. The water could squirt out up to a height that is half the depth of the pipe. Note that this doesn't generate a perpetual motion machine, since the "extra energy" available to squirt water out of the top was due to the fact that "on average" the water in the pipe only had to go halfway. Now that the pipe is full, no further net work is done on the pipe, so the amount of water that can be raised to this height is limited.

We can also figure out what happens if the pipe continues above the water surface. In that case, we need to take into account the entire body of water in the pipe for the motion. I will leave that as an exercise for you.

Finally, if you add friction, the math becomes more complicated, and depends on many factors. Bottom line - with enough friction, the water may lose so much energy when it fills the pipe that it never goes above the level of the water outside.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.