How to Calculate Water Leakage in an Inverted Bottle Experiment?

[evelynhott]

Homework Statement

I need to do this experiment, but I'm not sure if I did the calculation part correctly

Relevant Equations

Punch a hole in the cap of a one and a 1.5l cylindrical glass bottle. Fill the bottles half with water, then screw the cap back on. Turn it upside down and measure how much water comes out. Also, measure the height of the water in the bottle when the leakage stops. Using the measurement results, we determined what the air pressure was when the experiment was taken.

V_c-volume of the water that came out
V_all-volume of the water before leakage
S-area of the glass bottle
h-bottles high
h₁-height of air column before leakage
h₂-height of the air column after leakage
h_v,1-height of the water column before leakage
h_v,2-height of the water column after leakage

(1)p_aV₁=pV₂
p_aSh₁=pSh₂
p=p_ah₁/h₂
(2)p+ρgh_v,2=p_a
p_ah₁/h₂+ρgh_v,2=p_a
p_a(h₁/h₂-1)=-ρgh_v,2
p_a=(ρgh_v,2)/(1-(h₁/h₂))

(3)V_c=(h₁-h₂)S ->h₂=(V_c+Sh₁)/S
(4)V_all=h_v,2S
(5)h₂=(V_c-V_all)/(S)+h
(6)h₁=h-h_v,1=h-(V_all/S)
I substitute:
p_a=(h_v,2gV_cρ-h_v,2V_allgρ+h_v,2gSh)/V_c

 

[phinds]

Your calculations are illegible. At the very least you need to print them out MUCH bigger. According to forum standards, you are supposed to TYPE them, not show handwritten chicken scratches.

 

[phinds]

Much better now that you have typed them in. Water pressure depends on depth so it seems to me the water would squirt out more strongly at first and then trail off, linearly, I think. I find your equations a bit opaque so can't tell if that's accounted for.

You need to show the drawings as well so that it is clear what your symbols represent.

 

[haruspex]

the water would squirt out more strongly at first and then trail off

Yes, but how does that affect the result? Time was not discussed.

(1)paV1=pV2

It would not be isothermal initially. In principle it would continue to drip for a while as the air regains temperature. Not sure how quick that would be.
Did you get a reasonable value for p_a?

 

[phinds]

Yes, but how does that affect the result? Time was not discussed.

That was my point.

 

[haruspex]

That was my point.

Let me rephrase that… why does the time taken matter in finding the air pressure?

 

[evelynhott]

Did you get a reasonable value for pa?

Unfortunately not, when I replaced the values I got 5088,43Pa as the result. However, I know that this experiment is thermal and mechanical. I couldn't imagine where else the heat would appear, if not at the isothermal process.

 

[phinds]

Let me rephrase that… why does the time taken matter in finding the air pressure?

because as the water depletes from the container at a certain rate, the pressure of the air remaining will change accordingly.

 

[erobz]

Unfortunately not, when I replaced the values I got 5088,43Pa as the result. However, I know that this experiment is thermal and mechanical. I couldn't imagine where else the heat would appear, if not at the isothermal process.

I think the isothermal process is what is sought here. I think is meant to be a very small hole and a slow leak.

Just say the height of the bottle is $L$. What's the initial volume of air in terms of $L$?

Lable the height of the water $x$. What is the volume of air as a function of $x$ (using $L$).

That pressure at the surface over $A$ must balance the weight of the water in terms of $x$ remaining in the bottle when the leak has stopped( assuming the hole $\ll A$).

 

[evelynhott]

What's the initial volume of air in terms of L?

V_air=A(L-x)
I'm not sure: p_surface=ρxA

 

[erobz]

V_air=A(L-x)

That’s the volume as a function of $x$. The initial volume of air at $P_{atm}$ just involves $L$.

I'm not sure: p_surface=ρxA

That’s just the mass. You need the weight. Apply Newton’s second law for the equilibrium condition on the water( as the free body).

 

[evelynhott]

Apply Newton’s second law

Then F=am, the a in this case is g? P_atm=ρxAg

 

[evelynhott]

The initial volume of air at P_atm just involves L.

If we say that we filled half of the bottle, V_air=A(L/2), or how?
And where can I use the isothermal process? Please help!

 

[erobz]

If we say that we filled half of the bottle, V_air=A(L/2), or how?

Thats good.

And where can I use the isothermal process? Please help!

First just start with the force balance on the water in the vertical direction.

What forces are acting on the water that is in the bottle (assume the hole is negligibly small in comparison to the cross section of the bottle)? When the water stops leaking is any of the water mass accelerating?

What does Newtons Second law then say about the sum of the forces acting on the system ( the water mass)?

 

[Steve4Physics]

Hi @evelynhott. Since no one has yet said so, welcome to PF!

A few points…

When you refer to ‘heights’ I assume you mean lengths. E.g. the final length of the air column is $h_2$.

In Post #1 you wrote $(3)~V_c=(h_1-h_2)S \implies h_2=(V_c+Sh_1)/S$.
But the RHS is wrong (algebra error).

Also, you wrote $(4)~V_{all}=h_{v,2}S$.
Did you mean that? Have you forgotten to add the volume of water collected?

Note that you are implicitly assuming the cross-sectional area (S) of the bottle is constant. But bottles generally get narrower towards the neck.

And avoid losing marks due to misuse of significant figures. So when you say (Post #7) that you got a pressure of “5088.43Pa” that’s inappropriate.

 

[evelynhott]

In Post #1 you wrote

I typed it wrongly: (3)Vc=(h₂-h₁)S ->h₂=(Vc+Sh₁)/S

Also, you wrote

I typed it also wrongly: (4)V_all=h_v,1S

Thank you!

 

[evelynhott]

What forces are acting on the water that is in the bottle (assume the hole is negligibly small in comparison to the cross section of the bottle)? When the water stops leaking is any of the water mass accelerating?

At this point, the water mass is no longer accelerating. The leakage stops when the upward buoyant force equals the downward gravitational force. Is it right?
According to Newton's second law, the sum of forces acting on the water mass when it is not accelerating is zero.
ΣF = Fb - Fg = 0
And then Fb=Fg

 

[erobz]

At this point, the water mass is no longer accelerating. The leakage stops when the upward buoyant force equals the downward gravitational force. Is it right?
According to Newton's second law, the sum of forces acting on the water mass when it is not accelerating is zero.
ΣF = Fb - Fg = 0
And then Fb=Fg

Its not a buoyant force. There is a pressure (relative vacuum in the expanded gas ) acting above the upper surface of the water that is "pulling it up". Otherwise the equation is correct.

 

[evelynhott]

Its not a buoyant force. There is a pressure (relative vacuum in the expanded gas ) acting above the upper surface of the water that is "pulling it up". Otherwise the equation is correct.

Is viscosity ignored in this case?

And when the water stops leaking, it is no longer accelerating. The acceleration a is zero, and the sum of the forces acting on the water must also be zero. This means that the downward gravitational force on the water must be exactly balanced by the upward force exerted by the air pressure inside the bottle. Or how should I do it?

 

[erobz]

Is viscosity ignored in this case?

As far as an idealization goes, yes.

 

[erobz]

And when the water stops leaking, it is no longer accelerating. The acceleration a is zero, and the sum of the forces acting on the water must also be zero. This means that the downward gravitational force on the water must be exactly balanced by the upward force exerted by the air pressure inside the bottle. Or how should I do it?

Yeah, uniformly distributed pressure $P(x)$ acting over $A$ balances the weight of the water $W(x)$ when the water inside comes to rest. $P$ and $W$ are both functions of $x$ ( the height of the water in the container when equilibrium is reached).

Then $P(x)$ comes from isothermal expansion from the initial volume.

 

[Lnewqban]

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

Welcome, Evelyn!

What is the reasoning behind this equation, in which volumes and pressures are multiplied?
Why is Vc relevant?
Is the orifice submerged into the mass of water that leaked out?

When you write "we determined what the air pressure was when the experiment was taken", are you referring to the gas pressure inside the glass bottle after water has liked out?

 

[evelynhott]

What is the reasoning behind this equation, in which volumes and pressures are multiplied?
Why is Vc relevant?

I tried to substitute back the previously defined equations. But I see that I messed something up.
I tried to include Vc in the equations because it was written in the text of the task that we would also measure it.

Is the orifice submerged into the mass of water that leaked out?

I held the bottle so that the orifice didn't submerge.

 

[Lnewqban]

I held the bottle so that the orifice didn't submerge.

Is this diagram correct?

 

[evelynhott]

Is this diagram correct?

View attachment 341941

Yes, it's perfect

 

[Lnewqban]

Yes, it's perfect

Then, let's go to post #11 above.
While the mass of liquid is moving downwards, Newton's laws are in display, because the vertical forces acting in that mass are not balanced.

Therefore, after the natural leak stops, there is balance of vertical forces.
Static pressures directly above and below the orifice are equal.
Could you explain the reason for that to be?

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-1-fluids-density-and-pressure/

 

[evelynhott]

Then, let's go to post #11 above.
While the mass of liquid is moving downwards, Newton's laws are in display, because the vertical forces acting in that mass are not balanced.

Therefore, after the natural leak stops, there is balance of vertical forces.
Static pressures directly above and below the orifice are equal.
Could you explain the reason for that to be?

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-1-fluids-density-and-pressure/

I think the reason for the static pressures above and below the orifice being equal after the leak stops is due to Pascal's law.
The pressure difference causes the water to flow out.
As the water flows out, the height of the water column decreases. Eventually, the water stops flowing out, indicating that the pressure above and below the orifice has become equal. This occurs because the weight of the remaining water column above the orifice exerts a downward force, balancing the upward pressure exerted by the water below the orifice.

 

[Steve4Physics]

Is viscosity ignored in this case?

It might be worth noting that viscous forces occur only when there is fluid motion

In this problem, only the initial and final states need to be considered - in both of these states there is no fluid motion.

 

[Lnewqban]

I think the reason for the static pressures above and below the orifice being equal after the leak stops is due to Pascal's law.
The pressure difference causes the water to flow out.
As the water flows out, the height of the water column decreases. Eventually, the water stops flowing out, indicating that the pressure above and below the orifice has become equal. This occurs because the weight of the remaining water column above the orifice exerts a downward force, balancing the upward pressure exerted by the water below the orifice.

Pretty close; you are almost ready to redo the calculations of that pesky Pa.
Please, see how the initial atmospheric pressure above the liquid becomes Pa:

 

[haruspex]

because as the water depletes from the container at a certain rate, the pressure of the air remaining will change accordingly.

As the water depletes from the container at a certain rate, the pressure of the air remaining will change at a corresponding rate. But we do not care about the rate of either; provided it takes long enough/we wait long enough for the temperature to recover, PV will not have changed.

 

[erobz]

It might be worth noting that viscous forces occur only when there is fluid motion

In this problem, only the initial and final states need to be considered - in both of these states there is no fluid motion.

I think surface tension falls into the viscous effects category, and would be present in the static case. It wasn’t my intent to try to account for it though.

Anyhow, I don’t seem to be getting through to the OP, so I’ll step aside.

 

[Steve4Physics]

I think surface tension falls into the viscous effects category, and would be present in the static case.

To the best of my knowledge, surface tension is not considered to be a viscous force.

But you are correct - adhesive and surface tension forces will both be present. If they are ignored (which is probably intended) this introduces a systematic error.

It would be very interesting to repeat the experiment with a few drops of washing-up liquid added to the water. But I'm not sure if the experiment would then work very well!

 

[erobz]

To the best of my knowledge, surface tension is not considered to be a viscous force.

https://physics.stackexchange.com/questions/700620/change-in-surface-tension-of-inviscid-flow#:~:text=If we have a stationary fluid assumed inviscid,it possess' no surface tension (zero surface tension).

I've never studied it myself. But from this and other search results surface tension appears to be dependent on viscosity.

 

[evelynhott]

Pretty close; you are almost ready to redo the calculations of that pesky Pa.
Please, see how the initial atmospheric pressure above the liquid becomes Pa:

View attachment 341949

View attachment 341950

Very nice illustrations, thank you!
Now I'm a little confused. P_atm=P_a+ρh_v,1g?

 

[Steve4Physics]

https://physics.stackexchange.com/questions/700620/change-in-surface-tension-of-inviscid-flow#:~:text=If we have a stationary fluid assumed inviscid,it possess' no surface tension (zero surface tension).

I've never studied it myself. But from this and other search results surface tension appears to be dependent on viscosity.

They (viscosity and surface tension) both depend on inter-molecular forces - so they have a shared origin. But they have different mechanisms.

Since we're referencing physics.stackexchange (!):

"Both viscosity and surface tension are connected theoretically to inter-molecular forces, but they are still very different concepts."​

https://physics.stackexchange.com/questions/148792/viscosity-and-surface-tension

 

[haruspex]

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

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.

 

[Lnewqban]

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

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).

 

[evelynhott]

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

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

 

[evelynhott]

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!

 

[haruspex]

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.

 

[erobz]

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?

 

[haruspex]

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.

 

[Lnewqban]

So does the P_atm 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/

 

[haruspex]

@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]

@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.

 

[evelynhott]

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.

 

[erobz]

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 $x_{eq}$".

 

[Lnewqban]

@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.

 

[evelynhott]

that my explanations have not been confusing.

Not at all, thank you all for your help!

 

[haruspex]

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.