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

  • #36
evelynhott said:
pa=(hv,2gVcρ-hv,2Vallgρ+hv,2gSh)/Vc
That is dimensionally inconsistent. You have dropped a factor ##\rho## in the last term in the parentheses.
Other than that, I get the same expression by another route.

Perhaps that omission is just another typo. If so, please post your measured values.
 
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  • #37
evelynhott said:
Now I'm a little confused. Patm=Pa+ρhv,1g?
No, when we have the original volume of liquid still inside the glass bottle, which corresponds to hv,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).

Glass bottle-water-internal pressure pre-initial conditions.jpg
 
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  • #38
Lnewqban said:
No, when we have the original volume of liquid still inside the glass bottle, which corresponds to hv,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
So does the Patm push the surface of the water in the same way as the water pushes the cap?
 
  • #39
haruspex said:
If so, please post your measured values.
I will do the measurement again since I did it with a plastic bottle (I didn't have a glass bottle, but I already have one). It may cause inaccurate results. However, I will post the results. Thank you!
 
  • #40
erobz said:
surface tension appears to be dependent on viscosity.
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.
 
  • #41
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.
Is there surface tension also between the walls of the cylinder and the water?
 
  • #42
erobz said:
Is there surface tension also between the walls of the cylinder and the water?
Yes, but that is far less significant in terms of pressure because of the greater area.
 
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  • #43
evelynhott said:
So does the Patm push the surface of the water in the same way as the water pushes the cap?
Once the bottle is inverted (but still capped, as shown in the last diagram), the interior face of the cap holds a greater pressure than the atmospheric pressure acting on the exterior face of the cap.

The magnitude of that greater pressure equals the summation of both internal pressures: pressure of gas above water and pressure of column of water between surface and cap.

Please, see:
https://courses.lumenlearning.com/s...-variation-of-pressure-with-depth-in-a-fluid/

Hydrostatic-Pressure.png
 
  • #44
@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.
 
  • #45
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.
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.
 
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  • #46
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.
Thank you very much, I think the problem was caused by the fact that english is not my native language and I may have misunderstood something.
So in this case ##P(x_{eq})## stands for air pressure.
 
  • #47
evelynhott said:
So in this case ##P(x_{eq})## stands for air pressure.
The pressure inside the cylinder (gas/air) when the system comes to equilibrium position ##x_{eq}##. Its reads "The pressure as a function of ##x_{eq}##".
 
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  • #48
haruspex said:
@Lnewqban , you post lovely diagrams, but I am not sure where you are going with this.
@haruspex , have you noted anything incorrect or misleading in my posts?

Those diagrams have been my way to explain what I believe is applicable to this experiment.

I only hope that the OP (@evelynhott ) understood where we were going with those, and that my explanations have not been confusing.
 
  • #49
Lnewqban said:
that my explanations have not been confusing.
Not at all, thank you all for your help!
 
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  • #50
Lnewqban said:
have you noted anything incorrect or misleading in my posts?
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.
 
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  • #51
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.
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.
 
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  • #52
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.
Don’t be too hard on yourself. I had a hand in this by doing exactly what Haruspex said also. I was too eager to help that I skipped digesting whether or not the OP had it. In my mind I saw too many variables and I wanted to reduce them without really looking. That started the wild goose chase.
 
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  • #53
Hey @Lnewqban. Go easy on yourself. You’ve produced plenty of really helpful stuff. None of us are perfect so we get things wrong sometimes. (I speak with much experience of getting things wrong!)
 
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  • #54
Lnewqban said:
have miserabily failed to be of any help.
No, that is going too far. You have often been of real help on threads, and your diagrams are excellent.
Just try to follow my approach in post #50: take into account what exactly the OP is asking and where they appear stuck.
 
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  • #55
evelynhott said:
I calculated with the following data and got the following:
hv,2=0,11m , Vc=11,65ml, S=5,215·10^-3m^2 , h=32cm, patm= 105880Pa
I can see some possible problems with accuracy.

For clarity, l'll rewrite your equation as ##p_{atm}=h_{v,2}g\rho(Sh-V_{all}+V_c)/V_c##.
Your equation subtracts ##V_{all}## from Sh to get the original air volume then adds ##V_c## to get the final air volume.
You have used S and h as though the total volume of the bottle is their product, Sh. But your h is the entire bottle length, so includes narrower sections. With those numbers, Sh= 1.667L, about 10% too much.
But you do not say how you determined S, so maybe the bottle volume really is that large and you found S by dividing it by h?
And you don't state the value of ##V_{all}##. Presumably ##V_{all}=0.75L##, but how precisely? And is the total volume of the bottle really 1.5L, or is that just what it says on the label?

The nature of this experimental method is that it requires considerable accuracy in all these numbers to get a reasonable result.
 
  • #56
evelynhott said:
Vall=0,5L
Ok, but what about my other questions?
haruspex said:
you do not say how you determined S

haruspex said:
is the total volume of the bottle really 1.5L, or is that just what it says on the label?
 

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