What variables affect the height of a Heron's fountain?

In summary, the homework states that the height of the heron's fountain is directly dependent on pressure in "bottle b" and that the relevant parameters are V (volume of the liquid) ρ (viscosity of the liquid) and Δh (difference in heights - the length of a tube connecting "bottle a" and "bottle c").
  • #36
Thanks for that height-depth note, I was getting a bit tangled in this.

As for the pressure difference, the bigger the difference between in and outside the tube, the bigger the fountain would be.
And it can be expressed ΔP = h3ρg-(h2-h1)ρg - h1ρg = (h3-h2-2h1)ρg??
This looks so wrong... so I'll bet there must be something I missed in this calculation?
 
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  • #37
Dami121 said:
h3ρg-(h2-h1)ρg - h1ρg = (h3-h2-2h1)ρg??
Try that step again.
 
  • #38
Oh, stupid mistake,
ΔP=(h3-h2)ρg

Is that it?
 
  • #39
Dami121 said:
Oh, stupid mistake,
ΔP=(h3-h2)ρg

Is that it?
Yes.
 
  • #40
Well, it's been a pleasure and honor sir!
I think I have all I need. It took me ages , but now I understand it very well.
Thanks a lot :)
 
  • #41
@Physicist1011 (who asked me about this thread): do you understand it down to finding the pressure difference at post #38?
 
  • #42
No I don't. I don't understand how the differences in heights of the water affect the water height and I also don't understand how you got that equation.

Edit: or does the heights of the water affect the water fountain not only the differences between the heights of the water in the containers?
 
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  • #43
Physicist1011 said:
No I don't. I don't understand how the differences in heights of the water affect the water height and I also don't understand how you got that equation.
First thing is to understand how the pressures in the different airspaces are related.
If two airspaces are connected by a tube of air, what can you say about the two pressures?
 
  • #44
The pressures are the same? but why?
 
  • #45
Physicist1011 said:
The pressures are the same? but why?
If they were different, what would the air do?
 
  • #46
Move towards lower air pressure.
 
  • #47
Physicist1011 said:
Move towards lower air pressure.
Right.
Early in this thread I advised Dami to consider the set-up with a finger on top of the top tube so that no fountain occurs. This is so that everything is static, which makes the analysis easier.
So we can assume no air is flowing, so the pressures are the same. (When the fountain is flowing there will be a small difference in the air pressures.)

Next, consider a tube that is filled with water, connecting two reservoirs. For simplicity, we can just think about the pressures at the surfaces of the reservoirs.
What can you say about those two pressures? How are they related?
 
  • #48
I am not sure.
 
  • #49
Physicist1011 said:
I am not sure.
If two reservoirs are connected by a filled pipe they are effectively a single reservoir. What relates the pressures at two points in a reservoir?
 
  • #50
Sorry I am really not sure as this is what I am confused about.
How does the pressure at d relate to the difference in heights since the pressure is what relates to the fountain's height right?
 
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  • #51
Physicist1011 said:
Sorry I am really not sure as this is what I am confused about.
How does the pressure at d relate to the difference in heights since the pressure is what relates to the fountain's height right?
Consider a tank of water. How does the pressure at height y1 from the bottom relate to the pressure at height y2 from the bottom?
 
  • #52
These 2 heights of water are connected in a closed space (except for the water surface at the top). Such that the pressure of the lower height will be larger than that of the higher water height.
 
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  • #53
Physicist1011 said:
These 2 heights of water are connected in a closed space (except for the water surface at the top). Such that the pressure of the lower height will be larger than that of the higher water height.
Yes, but by how much?
 
  • #54
Pressure difference = pgh2(higher water height) - pgh1(lower water height)

Edit: oh so change in pressure from upper height to lower height = change in height *pg
 
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  • #55
Physicist1011 said:
Pressure difference = pgh2(higher water height) - pgh1(lower water height)

Edit: oh so change in pressure from upper height to lower height = change in height *pg
Right.
There are four locations to consider: the surface of the water in each container, and the base of the fountain. (By "base of fountain" I mean the point inside the top tube that is the same height as the surface of the top reservoir.)
If you can move from one such location to another only moving through air then you know the pressures at those locations are the same; if you can move from one to another only moving through water then you can relate the pressures by the height differences.
And don't forget, you have a finger on top of the top tube.

What equations can you write relating these four pressures?
 
  • #56
Pressure difference between B and A = pg*height difference between water surface B and A
Pressure difference between B and C - same
Pressure difference between C and A = pg*height difference between C and A
I am getting a little confused - is this correct?
 
  • #57
Physicist1011 said:
Pressure difference between B and A = pg*height difference between water surface B and A
Is there a path between those two surfaces that is entirely in water?
 
  • #58
Yes there is.. that's why there is a pressure difference.
 
  • #59
Physicist1011 said:
Yes there is.. that's why there is a pressure difference.
I am not seeing it.
Tube f (water) connects A and C.
Tube e (air) connects B and C.
Tube d (water) connects fountain base with B.
 
  • #60
Oh the pressure at A is atmospheric pressure so the difference in pressure between B and A is PB=Patm
 
  • #61
Physicist1011 said:
Oh the pressure at A is atmospheric pressure so the difference in pressure between B and A is PB=Patm
Atmospheric pressure is a background common to all points. We only need concern ourselves with gauge pressure, i.e. the excess above atmospheric.
But you still have not answered my question about A and B. What path is there between those two surfaces which either only passes through water or only passes through air?
 
  • #62
Well the water in B passes through a tube of water and comes out into air. So neither of those things you mentioned above happen because the water passes through water (in B) and then air (coming out of the tube d).
 
  • #63
Physicist1011 said:
Well the water in B passes through a tube of water and comes out into air. So neither of those things you mentioned above happen because the water passes through water (in B) and then air (coming out of the tube d).
As I reminded you, we are considering a static arrangement. Your finger is stopping the fountain, so nothing is actually passing anywhere.
My questions are in relation to paths that exist within this static arrangement. Could you thread a path from the one surface to the other without changing medium?
 
  • #64
Yes, the water is flowing in the tube to the 2nd surface so the medium does not change since the path is the same medium on both surfaces.
 
  • #65
Physicist1011 said:
Yes, the water is flowing in the tube to the 2nd surface so the medium does not change since the path is the same medium on both surfaces.
Which tube? They are labelled.
 
  • #66
tube d
 
  • #67
Physicist1011 said:
tube d
As I wrote in post #59, tube d connects reservoir B to the base of the fountain. It does not connect with reservoir A. There is no water path from the surface of reservoir B to the surface of reservoir A.
 
  • #68
Oh, tube e - tube of air connecting to 2 air mediums.
 
  • #69
Physicist1011 said:
Oh, tube e - tube of air connecting to 2 air mediums.
That connects the airspaces at B and C, not A and B..
 
  • #70
ok. tube f is a tube full of water which connects a water medium to another water medium.
 

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