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

  • #1
evelynhott
16
4
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.
431597886_3130771090389826_8017646907259626576_n.jpg

Vc-volume of the water that came out
Vall-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
hv,1-height of the water column before leakage
hv,2-height of the water column after leakage

(1)paV1=pV2
paSh1=pSh2
p=pah1/h2
(2)p+ρghv,2=pa
pah1/h2+ρghv,2=pa
pa(h1/h2-1)=-ρghv,2
pa=(ρghv,2)/(1-(h1/h2))

(3)Vc=(h1-h2)S ->h2=(Vc+Sh1)/S
(4)Vall=hv,2S
(5)h2=(Vc-Vall)/(S)+h
(6)h1=h-hv,1=h-(Vall/S)
I substitute:
pa=(hv,2gVcρ-hv,2Vallgρ+hv,2gSh)/Vc
 
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  • #2
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.
 
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  • #3
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.
 
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  • #4
phinds said:
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.
evelynhott said:
(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 pa?
 
  • #5
haruspex said:
Yes, but how does that affect the result? Time was not discussed.
That was my point.
 
  • #6
phinds said:
That was my point.
Let me rephrase that… why does the time taken matter in finding the air pressure?
 
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  • #7
haruspex said:
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.
 
  • #8
haruspex said:
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.
 
  • #9
evelynhott said:
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##).
 
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  • #10
erobz said:
What's the initial volume of air in terms of L?
Vair=A(L-x)
I'm not sure: psurface=ρxA
 
  • #11
evelynhott said:
Vair=A(L-x)
That’s the volume as a function of ##x##. The initial volume of air at ##P_{atm}## just involves ##L##.
evelynhott said:
I'm not sure: psurface=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).
 
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  • #12
erobz said:
Apply Newton’s second law
Then F=am, the a in this case is g? Patm=ρxAg
 
  • #13
erobz said:
The initial volume of air at Patm just involves L.
If we say that we filled half of the bottle, Vair=A(L/2), or how?
And where can I use the isothermal process? Please help!
 
  • #14
evelynhott said:
If we say that we filled half of the bottle, Vair=A(L/2), or how?
Thats good.
evelynhott said:
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)?
 
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  • #15
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.
 
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  • #16
Steve4Physics said:
In Post #1 you wrote
I typed it wrongly: (3)Vc=(h2-h1)S ->h2=(Vc+Sh1)/S
Steve4Physics said:
Also, you wrote
I typed it also wrongly: (4)Vall=hv,1S

Thank you!
 
  • #17
erobz said:
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
 
  • #18
evelynhott said:
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.
 
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  • #19
erobz said:
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?
 
  • #20
evelynhott said:
Is viscosity ignored in this case?
As far as an idealization goes, yes.
 
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  • #21
evelynhott said:
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.
 
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  • #22
evelynhott said:
I substitute:
pa=(hv,2gVcρ-hv,2Vallgρ+hv,2gSh)/Vc
Welcome, Evelyn! :smile:

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?
 
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  • #23
Lnewqban said:
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.
Lnewqban said:
Is the orifice submerged into the mass of water that leaked out?
I held the bottle so that the orifice didn't submerge.
 
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  • #24
evelynhott said:
I held the bottle so that the orifice didn't submerge.
Is this diagram correct?

Glass bottle-water-internal pressure.jpg
 
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  • #25
  • #26
evelynhott said:
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/
:cool:
 
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  • #27
Lnewqban said:
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/
:cool:
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.
 
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  • #28
evelynhott said:
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.
 
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  • #29
evelynhott said:
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:

Glass bottle-water-internal pressure initial conditions.jpg


Glass bottle-water-internal pressure final conditions.jpg
 
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  • #30
phinds said:
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.
 
  • #31
Steve4Physics said:
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.
 
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  • #32
erobz said:
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!
 
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  • #34
Lnewqban said:
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. Patm=Pa+ρhv,1g?
 
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  • #35
erobz said:
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
 
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