Please help with this physics experiment -- Water leaking out of inverted water bottle

evelynhott said:

p_a=(h_v,2gV_cρ-h_v,2V_allgρ+h_v,2gSh)/V_c

evelynhott said:

Now I'm a little confused. P_atm=P_a+ρh_v,1g?

Lnewqban said:

No, when we have the original volume of liquid still inside the glass bottle, which corresponds to h_v,1, only atmospheric pressure can be above the liquid.

At that initial point, we have done nothing to increase or reduce that pressure (no one drop has leaked out yet).

View attachment 341957

haruspex said:

If so, please post your measured values.

erobz said:

surface tension appears to be dependent on viscosity.

haruspex said:

That may be, but in terms of consequences, viscosity affects flows, not static arrangements. How it went from initial state to final here is of no interest.

As for the effect of surface tension here, if the hole has radius r then the vertical force force is at most ##2\pi\gamma r##, giving a pressure differential of ##2\frac\gamma r##.
With ##\gamma=73\cdot 10^{-3}N/m## and a radius of 1mm that gives ##146N/m^2##, or 1.5 mbar.

erobz said:

Is there surface tension also between the walls of the cylinder and the water?

evelynhott said:

So does the P_atm push the surface of the water in the same way as the water pushes the cap?

haruspex said:

@Lnewqban , you post lovely diagrams, but I am not sure where you are going with this.
Post #1 had one typo as @Steve4Physics pointed out, and what is either a second typo or a major error (probably accounting for the wildly incorrect result) noted in post #36.
Other than that, the equations in post #1 look right, so @evelynhott appears to understand all the principles. Indications to the contrary in later posts can be explained by simple misunderstandings in communication in both directions.

erobz said:

Well, if that's the case I'm just going to present what I was working them toward.

Isothermal Expansion:

$$P_{atm} \cancel{A} \frac{L}{2} = P(x_{eq}) \cancel{A} ( L-x_{eq})$$

$$ \implies P(x_{eq}) = P_{atm} \frac{L}{2 (L-x_{eq})} \tag{1}$$

Newtons Second:

$$ P(x_{eq}) A - W(x_{eq}) = 0 $$

$$P(x_{eq}) \cancel{A} = \rho \cancel{A} g x_{eq} \tag{2}$$

Sub ## 2 \to 1 ## for ##P(x_{eq})##

$$P_{atm} \frac{L}{2 (L-x_{eq})} = \rho g x_{eq} $$

Solve quadratic for ##x_{eq}## giving the equilibrium height.

Sub solution for ##x_{eq} \to 1## ( or ##2##) for ##P(x_{eq})##.

As far as the experiment goes you are measuring ##x_{eq}##, so we don't need to solve for it; it's meant for the purpose of theoretical verification.

evelynhott said:

So in this case ##P(x_{eq})## stands for air pressure.

haruspex said:

@Lnewqban , you post lovely diagrams, but I am not sure where you are going with this.

Lnewqban said:

that my explanations have not been confusing.

Lnewqban said:

have you noted anything incorrect or misleading in my posts?

haruspex said:

No, but neither have I seen much in them which addresses the question asked, namely, where has the OP gone wrong?

The first step should be to read where the OP got to and check whether there are any errors so far. In this case, the equation obtained was substantially correct and, if completely correct, should have been adequate for the OP's purposes. That being so, there does not appear to be any merit in going back to basics, as in post #26.

Of the 47 posts, including this one, I see 17 by the OP and less than 10 others that make a reasonable attempt to address that question. Sadly, this is not unusual.

Lnewqban said:

I see.
Thank you for your honest opinion, haruspex.

Evidently, my good intentions, not being supported by the deep understanding of these matters that you and other prominent members enjoy, have miserabily failed to be of any help.

My sincere apologies to all who may have been confused or annoyed by my past posts and diagrams, in this and many other PF threads, during a few, indeed enjoyable for me, years of learning and sharing.

Lnewqban said:

have miserabily failed to be of any help.