Physically speaking, what happens in Torricelli’s experiment?

  • #1
adjurovich
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I understand what happens in experiment, but my understanding of pressures isn’t the perfect.

Torricelli used a one meter long glass tube with opening on one end with mercury inside it. He then dipped the tube into a bowl of mercury and mercury started to dip from tube. However it stopped after some time…

Now, physically speaking what actually happens there? Why does mercury leak into the bowl at all? The only explanation I could think of is that hydrostatic pressure inside the tube is higher than atmospheric pressure (acting on mercury in bowl) and thus gravity causes mercury to push on the mercury inside the bowl — just like piston would. So the mercury level raises.

However, why does it stop when hydrostatic pressure of mercury inside the tube becomes equal to the atmospheric pressure?
 
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  • #2
adjurovich said:
I understand what happens in experiment, but my understanding of pressures isn’t the perfect.

Torricelli used a one meter long glass tube with opening on one end with mercury inside it. He then dipped the tube into a bowl of mercury and mercury started to dip from tube. However it stopped after some time…

Now, physically speaking what actually happens there? Why does mercury leak into the bowl at all? The only explanation I could think of is that hydrostatic pressure inside the tube is higher than atmospheric pressure (acting on mercury in bowl) and thus gravity causes mercury to push on the mercury inside the bowl — just like piston would. So the mercury level raises.

However, why does it stop when hydrostatic pressure of mercury inside the tube becomes equal to the atmospheric pressure?
1713467754095.png
Think of the force pushing the mercury up into the tube and the force pulling it down.
I think it's simpler with the situation in the picture above.
 
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  • #3
adjurovich said:
... However, why does it stop when hydrostatic pressure of mercury inside the tube becomes equal to the atmospheric pressure?
It does not suddenly stop, it oscillates up and down for a short time around the level of balance of forces.
Why would it not stop?
 
  • #4
Lnewqban said:
It does not suddenly stop, it oscillates up and down for a short time around the level of balance of forces.
Why would it not stop?
Oscillating and time interval does not really matter here, we are talking about basic fluid statics for high school. It of course does stop as one would expect it, but what causes it to stop?
 
  • #5
Philip Koeck said:
View attachment 343661Think of the force pushing the mercury up into the tube and the force pulling it down.
I think it's simpler with the situation in the picture above.
Before placing the tube into the bowl (basin, or whatever), the top of the fluid is exposed to atmospheric pressure. When we put it inside (since we consider the tube to be full), the mercury at the bottom of the tube will displace some mercury in the bowl, and now the opening of the tube will be a little submerged. According to experiment, the mercury stops to exit the tube when enough of it has left the tube so that the hydrostatic pressure at the bottom of the tube (where opening is) is equal to the atmospheric pressure. However, pressure underneath the tube opening is now equal to atmospheric + hydrostatic (since the opening is a submerged). If we take in account this small hydrostatic pressure, then equilibrium is achieved when the pressure inside of the tube is equal to the pressure underneath it (atmospheric + small hydrostatic pressure), but not only atmospheric. This is how I understood it. Please correct me where I am wrong
 
  • #6
adjurovich said:
Oscillating and time interval does not really matter here, we are talking about basic fluid statics for high school. It of course does stop as one would expect it, but what causes it to stop?
That is why I mentioned the oscillation: there is a force bringing the fluid inside the tube back to a unique level of equilibrium, once it overshoots it in both directions.

The fluid inside the tube has a weight.
Above it, there is vacuum.
Below it, another mass of fluid to be lifted or lowered.

As the vacuum increases, the weight of the fluid inside the tube gets reduced, and the mass of fluid below increases, and vice-verse.
 
  • #7
adjurovich said:
Before placing the tube into the bowl (basin, or whatever), the top of the fluid is exposed to atmospheric pressure. When we put it inside (since we consider the tube to be full), the mercury at the bottom of the tube will displace some mercury in the bowl, and now the opening of the tube will be a little submerged. According to experiment, the mercury stops to exit the tube when enough of it has left the tube so that the hydrostatic pressure at the bottom of the tube (where opening is) is equal to the atmospheric pressure. However, pressure underneath the tube opening is now equal to atmospheric + hydrostatic (since the opening is a submerged). If we take in account this small hydrostatic pressure, then equilibrium is achieved when the pressure inside of the tube is equal to the pressure underneath it (atmospheric + small hydrostatic pressure), but not only atmospheric. This is how I understood it. Please correct me where I am wrong
I think it might be easier to discuss if you refer to the picture below since you can clearly see which forces are balanced. Give it a try.
1713505867923.png
 
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  • #8
Philip Koeck said:
I think it might be easier to discuss if you refer to the picture below since you can clearly see which forces are balanced. Give it a try.
View attachment 343682
This one seems pretty obvious. We can see that atmosphere is pushing on the left free surface, while at the same elevation on the right side, hydrostatic pressure (caused by the weight of the fluid) is pushing and balancing it.

Could we make some analogy to Torricelli’s experiment? Some of the fluid will leave the tube, and the remaining will be balanced by atmospheric pressure acting from below (I guess), because the hydrostatic pressure of the remaining fluid won’t be enough to “push” upwards fluid in the bowl. So it reaches equilibrium.

I added “I guess” because the “acting in all directions” aspect of pressure is confusing me at the times even though it does make sense to me perfectly in the most cases
 
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  • #9
adjurovich said:
This one seems pretty obvious. We can see that atmosphere is pushing on the left free surface, while at the same elevation on the right side, hydrostatic pressure (caused by the weight of the fluid) is pushing and balancing it.

Could we make some analogy to Torricelli’s experiment? Some of the fluid will leave the tube, and the remaining will be balanced by atmospheric pressure acting from below (I guess), because the hydrostatic pressure of the remaining fluid won’t be enough to “push” upwards fluid in the bowl. So it reaches equilibrium.

I added “I guess” because the “acting in all directions” aspect of pressure is confusing me at the times even though it does make sense to me perfectly in the most cases
I think you have it. With the bowl it's just a bit more complicated since the surface area of the mercury in the bowl is different from the cross-sectional area of the tube, but the balance of forces acting on the column of mercury in the tube is exactly the same.

It's true that pressure acts in all directions equally, but for the mercury in the bowl the atmospheric pressure can only push downwards, which means it pushes the mercury column in the tube upwards.
 
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  • #10
This last paragraph is exactly what confuses me. It does seem logical though if I think of it this way “if mercury in tube pushes downwards, the mercury in the bowl tends to go upwards because it will be “pushed”. But if atmosphere pushes it downwards, the mercury in the bowl can’t go upwards and equilibrium is achieved. It’s weird to think of two things pushing downwards and balancing each other, but at the same time works!

The only way it makes sense is if pressure exerted by fluid in tube spreads in all directions, including upwards, but atmospheric pressure acts only downwards. I think (but am not completely sure) the thing I mentioned above might be considered Pascal’s principle… but as I said I’m not sure because Pascal’s principle refers to fluid in closed system, and this one isn’t.
 
  • #11
adjurovich said:
It’s weird to think of two things pushing downwards and balancing each other, but at the same time works!
adjurovich said:
... but atmospheric pressure acts only downwards...
... and it makes a U-turn at the bottom end of the tube and pushes upwards, helped by the hydrostatic pressure that exists at that point.
Perfectly equivalent to the diagram shown in post #7 above.
 
  • #12
adjurovich said:
This last paragraph is exactly what confuses me. It does seem logical though if I think of it this way “if mercury in tube pushes downwards, the mercury in the bowl tends to go upwards because it will be “pushed”. But if atmosphere pushes it downwards, the mercury in the bowl can’t go upwards and equilibrium is achieved. It’s weird to think of two things pushing downwards and balancing each other, but at the same time works!

The only way it makes sense is if pressure exerted by fluid in tube spreads in all directions, including upwards, but atmospheric pressure acts only downwards. I think (but am not completely sure) the thing I mentioned above might be considered Pascal’s principle… but as I said I’m not sure because Pascal’s principle refers to fluid in closed system, and this one isn’t.
You need to do a free body diagram of the mercury column in the tube. From the free surface of the base to the underside of the tube you apply hydrostatic equation to find the pressure acting upward of the base of the column.

1713537995060.png



Sum the forces acting on the system under equilibrium:

$$ P_{atm} A - \rho g A h + P_v A = 0 $$

##P_v## is mercury vapor pressure (assuming tube was initially inverted without the introduction of air)
##A## is the cross-sectional area of the tube
## \rho## is the density of the mercury
 
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  • #13
erobz said:
You need to do a free body diagram of the mercury column in the tube. From the free surface of the base to the underside of the tube you apply hydrostatic equation to find the pressure acting upward of the base of the column.

View attachment 343704


Sum the forces acting on the system under equilibrium:

$$ P_{atm} A - \rho g A h + P_v A = 0 $$

##P_v## is mercury vapor pressure (assuming tube was initially inverted without the introduction of air)
##A## is the cross-sectional area of the tube
## \rho## is the density of the mercury
Thanks for effort. If I got it right, atmospheric pressure is acting in all directions inside the fluid? So that’s how it eventually balances hydrostatic pressure coming from tube:

(Sorry for poor illustration, I made it on my phone)

IMG_5690.jpeg


(Note: these aren’t vectors, these are just showing the direction of pressure).
 
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  • #14
adjurovich said:
Thanks for effort. If I got it right, atmospheric pressure is acting in all directions inside the fluid? So that’s how it eventually balances hydrostatic pressure coming from tube:

(Sorry for poor illustration, I made it on my phone)

View attachment 343721

(Note: these aren’t vectors, these are just showing the direction of pressure).
Yeah, it looks like you understand. That’s why I drew that line going from the surface back up to the elevation of that surface. Between the two endpoints there is no hydrostatic pressure difference. We know the pressure at one of the endpoints …we deduce the pressure at the other to be equivalent.
 
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  • #15
I
adjurovich said:
Thanks for effort. If I got it right, atmospheric pressure is acting in all directions inside the fluid? So that’s how it eventually balances hydrostatic pressure coming from tube:
...
I recommend you to visualize isobaric planes for static fluid problems.
Unless the fluid is in movement and/or accelerating (other than vertically, as shown in the diagram showing fluid located inside an accelerating elevator), those isobaric planes, which contain all points with similar static pressure, will always be perfectly horizontal.

accel_up.gif

Nd9GcR7lRzBdZxxL69VONFHO4hOyc_garL-9DN9Og&usqp=CAU.jpg

Nd9GcT7vDy5LxEhxfbIvgbY94Dg0UeOQXHIt4T42g&usqp=CAU.png

main-qimg-c94a1761e04a83f3e930191075fb9508.png

main-qimg-c785ebee37c9ecc58dfb9d5d9160bf8c-pjlq.jpg
 
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