Torricellian tube inclined

[Rugile]

Homework Statement

A Torricellian tube (one end open, one end closed) of 1m length is filled completely with mercury and situated vertically the closed end down. Then the tube is flipped around 180 degrees and there is 760mm of mercury left. If we incline the tube at an angle $\phi$, what is the height of column of mercury in the tube? Find the dependency h($\phi$), if when phi = 0 the tube is vertical, and when phi = 90 it is horizontal.

Homework Equations

$p = \rho g h$

The Attempt at a Solution

It is not said in the problem statement that the tube is emerged into a bowl of mercury, thus the height of the column when inclined can't stay 760mm. So I'm not really sure what happens when we rotate the tube - my guess is that some air goes through the mercury up to where Torricellian vacuum is supposed to be? If so, I have an equation (which is based on intuition only, can't really prove it) of equilibrium, but I don't really know if it is right: $p_a = p_{\phi} + \rho g h_{\phi}; p_a = p_a sin\phi + \rho g h_{\phi}$, p_a is atmospheric pressure. It does make sense in a way, that when phi = 0, sin(phi) = 0 and p_a = ρgh, and when phi = 90, then h = 0. But I'm unable to prove it with physics... Any hints?

 

[BvU]

It is not said in the problem statement that the tube is emerged into a bowl of mercury

You mean: It is not said in the problem statement that the bottom end of the tube is immersed into a bowl of mercury. But the exercise really wants you to make that assumption.

No air sneaking in either.

There is an analogy with communicating vessels (not identical, but Δp = ρgh in both cases). What happens when the third one from the left is tilted more and more (while remaining of the same length) ?

This Δp = ρgh is something you should take under thorough consideration: what does it mean, precisely. (note I already added a Δ).

Can't distill sense out of your equations. No idea what $p_{\phi}$ stands for.

(the way you tackle this made me think of an Einstein joke where the blackboard shows E = ma² crossed out, mb² crossed out too, and mc² marked with a big eureka exclamation point. Sorry).

Imagine the tube is upright and has a huge diameter. Then think of a thin cylinder at an angle $\phi$, well within the big tube. Thin cylinder is long enough to go from vacuum above to Hg bath at foot. Close top end of thin cylinder and let big tube drain. Would anything change for the thin tube ?By the way, I do find the exercise wording somewhat misleading: do they want the (average) length of the mercury column ? or the difference in height ?

 

[Rugile]

Thank you for the answer. Well if you say it is safe to assume that the bottom of the tube is emerged in mercury, then I do know how to solve it :)

My equation had the idea that p(phi) is the pressure of air in the tube above mercury, if there is any air sneaking. As I said, it was completely unmotivated, just intuition (no hard feelings for the joke, it is funny!).

They want the length of mercury column in the tube. Not difference, just dependence of how high the column is depending on the inclined angle.

 

[BvU]

Problem is height is in the vertical direction, length can be along the tube. Up to you. As soon as the top of the tube is less than 760 mm above the liquid level, there is no more vacuum.

 

[HalfLostAndSearching]

I would first clarify some details with the problem statement. Is the tube indeed submerged in a bowl of mercury? If so, what is the height of the bowl? Also, what is the initial pressure inside the tube before it is flipped around? These details are important in determining the pressure equilibrium inside the tube.

Assuming that the tube is indeed submerged in a bowl of mercury and the initial pressure inside the tube is atmospheric pressure, then your intuition is correct. When the tube is inclined, the air inside the tube will push the mercury up to create a new level of equilibrium, where the pressure at the top of the air column is equal to the atmospheric pressure plus the pressure due to the height of the mercury column. This can be expressed as:

p_a = p_{\phi} + \rho g h_{\phi}

Where p_a is the atmospheric pressure, p_{\phi} is the pressure due to the air column, \rho is the density of mercury, g is the acceleration due to gravity, and h_{\phi} is the height of the mercury column when the tube is inclined at an angle \phi.

To find the dependency h(\phi), we can rearrange the equation to solve for h_{\phi}:

h_{\phi} = \frac{p_a - p_{\phi}}{\rho g}

Since p_a and p_{\phi} are constants, we can also write this as:

h_{\phi} = \frac{\Delta p}{\rho g}

Where \Delta p is the difference in pressure between the atmospheric pressure and the pressure due to the air column. This equation shows that the height of the mercury column is directly proportional to the pressure difference and inversely proportional to the density of mercury and the acceleration due to gravity.

When \phi = 0, the tube is vertical and there is no air column, so the pressure due to the air column is zero and h_{\phi} = 0. When \phi = 90, the tube is horizontal and the pressure due to the air column is equal to the atmospheric pressure, so h_{\phi} = \frac{p_a}{\rho g}. This confirms your intuition that when \phi = 0, the tube is vertical and when \phi = 90, the tube is horizontal.

In summary, when the Torricellian tube is inclined, the height of the mercury column is given by