Gauge pressure of water oozing out of a pipe

[brotherbobby]

Homework Statement

Water oozes out slowly from a pipe doubles back under the opening (spout) for a considerable distance and is held there by atmospheric pressure as shown in the figure below, before it detaches and falls. Four points are marked at the top and bottom of the water layers, inside and outside the pipe. Rank those four points according to the $\mathbf{gauge\; pressure}$ of the water there, the most positive first.

Relevant Equations

Gauge pressure to a liquid at a "depth" $h$ from the surface open to air : $P_G = \rho_L gh$ where $\rho_L$ is the density of the liquid.

Assuming water to flow out of the pipe with the same speed as inside and the thickness of water column $h_{ab} = h_{cd} = h$, my answer would be $\mathbf{(P_b = P_c) > (P_a = P_d)}$.

My reasoning is as follows : at positions $a\; \text{and}\; d$ the gauge pressure is 0 and the total pressure is due to atmosphere ($P_{\text{atm}}$). Both the positions are on "top" of the column and open to air.

The gauge pressure at places $b\; \text{and}\; c$ is given by the formula above : $P_G = \rho_L gh$. As $h$ is the same for the two parts, the pressures are also the same.

Is my answer correct?

 

[haruspex]

Consider the forces acting on the water under the pipe. How do they balance? Write the actual equation.

 

[brotherbobby]

Consider the forces acting on the water under the pipe. How do they balance? Write the actual equation.

Yes clearly the water is in equilibrium before it falls off. It is moving uniformly so there can't be a net force in the $x$ direction. In the $y$ direction, gravitational force on the water $= - mg \hat y$ is balanced by the air and gauge pressure on the water in the $+ \hat y$ direction.

$\vec{F}_G + \vec {F}_{\text{air}} = 0$

 

[haruspex]

Yes clearly the water is in equilibrium before it falls off. It is moving uniformly so there can't be a net force in the $x$ direction. In the $y$ direction, gravitational force on the water $= - mg \hat y$ is balanced by the air and gauge pressure on the water in the $+ \hat y$ direction.

$\vec{F}_G + \vec {F}_{\text{air}} = 0$

So which is greater, Pc or Pd?

 

[brotherbobby]

So which is greater, Pc or Pd?

Force due to air $\vec F_{\text{air}}$ is acting up. Weight due to the liquid column $\vec F_G = m\vec g$ is acting down. This is however not the same to a beaker filled with water - air is below the liquid. Hence, $P_c > P_d$. (We note that both $d \; \text{and} \; a$ need to have the same pressure being connected to air. )

Is this right?

 

[haruspex]

View attachment 255859

Force due to air $\vec F_{\text{air}}$ is acting up. Weight due to the liquid column $\vec F_G = m\vec g$ is acting down. This is however not the same to a beaker filled with water - air is below the liquid. Hence, $P_c > P_d$. (We note that both $d \; \text{and} \; a$ need to have the same pressure being connected to air. )

Is this right?

I don't understand your reasoning.
mg acts down, Pc acts down, Pd = P_atm acts up. What does that tell you?

 

[brotherbobby]

I don't understand your reasoning.
mg acts down, Pc acts down, Pd = P_atm acts up. What does that tell you?

It tells me that, contrary to my reasoning, $P_d > P_c$.
So it means then that $P_c$ is lower than atmospheric pressure ($P_d = P_{\text{atm}}$).

Hence the answer to the (original) question in the book, which asked for the pressures at the four points to be arranged in descending order will read : $\color{blue}{P_b > P_a = P_d > P_c}$.

Is this the right answer?

 

[haruspex]

View attachment 255902

It tells me that, contrary to my reasoning, $P_d > P_c$.
So it means then that $P_c$ is lower than atmospheric pressure ($P_d = P_{\text{atm}}$).

Hence the answer to the (original) question in the book, which asked for the pressures at the four points to be arranged in descending order will read : $\color{blue}{P_b > P_a = P_d > P_c}$.

Is this the right answer?

I believe that is right.

 

[brotherbobby]

I believe that is right.

Thank you for your patience. The problem looked deceptively simple.

On a different but related point, how is it at all possible that water slides down the side of the tube for a while before falling? Does it have something to do with surface tension?
Or conversely, if indeed it could drag along the (lower) surface of the tube, why doesn't it carry all along that way to the end of the tube? Is there a way to actually calculate where it is along the tube that it would lose contact and fall under gravity?

Thank you a lot.

 

[gmax137]

I think the phenomenon is more complicated than simple pressure difference. lookup "teapot effect."

 

[haruspex]

Thank you for your patience. The problem looked deceptively simple.

On a different but related point, how is it at all possible that water slides down the side of the tube for a while before falling? Does it have something to do with surface tension?
Or conversely, if indeed it could drag along the (lower) surface of the tube, why doesn't it carry all along that way to the end of the tube? Is there a way to actually calculate where it is along the tube that it would lose contact and fall under gravity?

Thank you a lot.

Yes, it's to do with surface tension. Just as it can take a while for a drip to form from a tap, the water running back underneath can take a while to detach. As it runs along, it is gradually sagging.

 

[haruspex]

I think the phenomenon is more complicated than simple pressure difference. lookup "teapot effect."

Strictly speaking, the "teapot effect" refers to the way that the flow can separate into the stream that goes into the cup and a stream that stays attached to the spout.
That it can stay attached at all is down to air pressure and surface tension, and is no different from a drip of water hanging from the underside of a surface until it gets too big.