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Height of Water in two conical tanks

1. Homework Statement
Two vertical tanks have their vertices connected by a short horizontal pipe. One tank initially full of water has an altitude of 1.8 meters and a diameter of base 2.2 meters. The other tank initially empty has an altitude 2.7 meters and a diameter of a base 2.4 meters. If the water is allowed to flow through the connecting pipe, find the level to which the water will ultimately rise in the empty tank.

2. Homework Equations
V=1/3(π/4)d2h

3. The Attempt at a Solution
I know for sure that the water that would go into the other tank would be one half of the volume of the full tank so
V=1/3(π/4)d2h
V=1/3(π/4)2.22*1.8
V=2.281m3 divided by two = 1.14m3 of water would go to the empty tank.

Please enlighten me on how could i possibly find the height of the other tank, i dont even know what the height of the full tank would be after the discharge so any help would be well appreciated.
 
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1. Homework Statement
Two vertical tanks have their vertices connected by a short horizontal pipe. One tank initially full of water has an altitude of 1.8 meters and a diameter of base 2.2 meters. The other tank initially empty has an altitude 2.7 meters and a diameter of a base 2.4 meters. If the water is allowed to flow through the connecting pipe, find the level to which the water will ultimately rise in the empty tank.

2. Homework Equations
V=1/3(π/4)d2h

3. The Attempt at a Solution
I know for sure that the water that would go into the other tank would be one half of the volume of the full tank
Oh really? What makes you so sure?

Chet
 

BvU

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I know for sure that the water that would go into the other tank would be one half of the volume of the full tank
And why would that be? Do tanks know about having to share equally or is there some other force that drives the water to flow ?

[edit] Drat, Oh, Chet is so fast !

He is someone pulling legs ? vertices connected ? conical tanks ?
 
Oh really? What makes you so sure?

Chet
Yeah I started doubting it too... if the heights are different there would be different pressures and with different pressures there would be different volumes. But the pipe is not even described I don't really have an idea what to do here. So please just help me.
 

BvU

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Indulge us and make a drawing ... perhaps these vertices are at the bottom and the bases are at the top ?

And the pipe can't really be that short, can it ? :smile:
 
And why would that be? Do tanks know about having to share equally or is there some other force that drives the water to flow ?

[edit] Drat, Oh, Chet is so fast !

He is someone pulling legs ? vertices connected ? conical tanks ?
What you see in my post is the whole problem I know its really a problem designed just to trip people
Indulge us and make a drawing ... perhaps these vertices are at the bottom and the bases are at the top ?

And the pipe can't really be that short, can it ? :smile:
That's what im wondering about too... I even thought there could be a wormhole connecting the two of them. But seriously there is no drawing or anything like that. What I typed is the whole problem itself.
 
There are actually many more problems like this so can somebody just enlighten me how to solve for the height of a partially filed cone with a given volume of water?
 
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The volume of the connecting pipe is negligible. If the final heights in the tanks are different then, as you said, the pressures at the two ends of the connecting pipe will be different. But they can't be different because the system would then not be in equilibrium. So, if h is the final height of liquid in the originally full tank, what is the final height of the liquid in the other tank? What was the original volume of liquid in the first tank if it was originally full? Does this total volume of liquid change when the system re-equilibrates?

Chet
 

SteamKing

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Yeah I started doubting it too... if the heights are different there would be different pressures and with different pressures there would be different volumes. But the pipe is not even described I don't really have an idea what to do here. So please just help me.
The bold portion of the quote above tells you what you need to know.

You also say, "with different pressures there would be different volumes". How do you know this? Have you studied Pascal's Law?

https://en.wikipedia.org/wiki/Pascal's_law

A more informal statement is, "Water seeks its own level."

The pipe is just a means to allow the water to flow from one tank to another. Further description of it is unnecessary to the solution.
 

BvU

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This ring a bell ?


800px-Communicating_vessels.svg.png
 

BvU

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I can draw fast, but this one was stolen -- oops, I hope I don't run into trouble over it ... (it appears to be clickable too ! :wink: )
 

BvU

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There are actually many more problems like this so can somebody just enlighten me how to solve for the height of a partially filed cone with a given volume of water?
Well, you already gave an expression (*). Let's take it all seriously and assume that the vertices are really at the bottom end and the bases up in the air.

Snag: d is now d(h) if h is not the full height any more ....
 
The bold portion of the quote above tells you what you need to know.

You also say, "with different pressures there would be different volumes". How do you know this? Have you studied Pascal's Law?

https://en.wikipedia.org/wiki/Pascal's_law

A more informal statement is, "Water seeks its own level."

The pipe is just a means to allow the water to flow from one tank to another. Further description of it is unnecessary to the solution.
No what I mean to say is that there would be different volumes per each tank such as the distribution wont be 50/50 between them because of the difference in height? Actually I'm not that even sure anymore I'm really just completely assuming here...
 

BvU

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Here's the drawing you should have made. To scale!

Conical tanks.jpg
 

SteamKing

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No what I mean to say is that there would be different volumes per each tank such as the distribution wont be 50/50 between them because of the difference in height? Actually I'm not that even sure anymore I'm really just completely assuming here...
That's why I pointed you to Pascal's Law, and BvU kindly posted the graphic in Post #10.

Both are saying the same thing: it's not the volume which causes pressure, it's the ?
 

BvU

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This is what my old paranoid self thought of in post #3
You know: base at the base and vertex at the top...

ROTFL they say nowadays :smile:

Conical tanks2.jpg
 
So I tried solving a similar problem to this one which is.
A piece of lead pipe of inner diameter 1 5/8 inch, outer diameter 2 1/8 inch and length 13 feet has been melted in an open conical pot radius 12 inches and altitude of 18 inches find the depth of the molten material.

Volume of pipe= π/4D2h
=π/4 (2.1252-1.6252)(13*12)
=229.73 in3

Pot
r=12in
h=18in

r/h=12/18
r=2/3h

Volume of cone
v=πr2h/3
The total volume inside the cone would be 229.73in3
229.73=π(2/3h)2h/3
h=7.903in would be the depth of the cone occupied by the molted lead

The answer in this question is supposed to be 8.81inches.... Is there something wrong with my computation or is it wrong overall? Also there have been cases where the answer I get is correct and the written in my reviewer is wrong so its possible that I'm correct... I just need confirmation if what I did was the right approach.
 

BvU

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If it helps you: I get the same result. Don't see any easy ratio in 8.81/7.903 or in its square or third power either, so no idea what went wrong with the book answer...
 
If it helps you: I get the same result. Don't see any easy ratio in 8.81/7.903 or in its square or third power either, so no idea what went wrong with the book answer...
Thank you for the confirmation,.
 
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In my view, when solving a physics problem, it is important to be able to articulate what is happening physically. The rest is just "doing the arithmetic."

So, in your original problem, there are two key concepts that need to be captured:
  1. The final height of liquid in the two tanks will be the same (you articulated this correctly)
  2. The total final volume of liquid in the two tanks is the same as the initial volume.
Everything else is "arithmetic."

Chet
 
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