Is it like this?
Hydrodynamics aren't really my thing, I hope that it's right.
Yes, but we do not need to fill in a number for atmospheric pressure. That is just a background value that will cancel out in the end. Call it Pa. So rewrite the equation using that and the variables we have defined for the three heights, A, B and C.
Okay so we have:
Pressure at A -
P = h1ρg + Pa = 0+Pa
Pressure at B -
P = h2ρg + Pa
Pressure at C -
P = h3ρg + Pa
How do you get that? What connects the contents of bottles B and C?
Of course that's not right, sorry, I have no idea how I got to that.
The bottles are connected with a tube (full of air)
The problem is that I really don't know what is the pressure force applied on the air.
What will happen if the pressures at B and C differ?
Oh I'm very sorry! I thought replied, and was wondering why you aren't writing back.
The pressures at B and C can't differ because that would mean the water would be pushed into the tube (with lower pressure).
Hence the pressure at B and C must be the same, that is P=h3ρg+Pa.
That is all, or no?
They are the same, but not for that reason. What tube connects B and C? What is in that tube?
Well the bottles are connected with a tube full of air, so I guess the pascal law comes in, saying that pressure applied on the air must be same in every point in the given connected medium.
There will be a very small pressure difference because of the weight of air in the tube, but we can ignore it.
If the tube were full of water then the pressures would be different (and the fountain would not work).
Edit: what I previously wrote in this next sentence was not what I meant to ask. See next post.
Next, what is the pressure inside tube d at A? Look at what connects this to B.
The tube is full of water, so I think the pressure in tube d at A would be
P = h1ρg+h3ρg ?
Of course the pressure changes there, but the initial pressure from the air in bottle B remains the same in the whole liquid no?
Is the pressure inside the tube d at A more or less than the pressure at B?
Not sure what you mean.
Remember we started by supposing you have your finger on top of the tube d to inhibit the fountain so that everything is static. In that arrangement, there are two rules you can apply:
If two points are connected by a path passing only through air then the pressures are near enough equal;
If two points are connected by a path passing only through water then the difference in pressures will be ρwgh, where h is the height difference between the two points.
Wow, I can't get my head around it now... I will go get some sleep and respond tommorow.
But let me try at least, since the height A is smaller than height B I guess the pressure in the tube d at A must be smaller than at B.
Then the pressure at A in the tube would be:
P = h3ρg-(h2-h1)ρg ?
To be clear, the way you are using h1 ... h3 they are depths from the top, not heights from the bottom. So you mean that the depth at A is less than the depth at B.
So what is the pressure difference between inside the tube and outside the tube at A?
How does that pressure difference relate to the height of the fountain?
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?
Try that step again.
Oh, stupid mistake,
Is that it?
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 :)
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