Fluid Mechanics: Weight vs Pressure

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Homework Help Overview

The problem involves a thin uniform circular tube in a vertical plane filled with two non-miscible liquids of different densities. The setup describes the equilibrium condition where the interface between the liquids makes a specific angle with the vertical, and the task is to determine the ratio of the densities of the liquids.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equilibrium condition and the role of weight versus pressure in the analysis. There are attempts to equate weights on both sides of the interface, and questions arise regarding the correct application of fluid mechanics principles.

Discussion Status

Some participants have provided guidance on using pressure differentials instead of weights for equilibrium analysis. There is an ongoing exploration of the assumptions made in the original approach, with some participants expressing confusion about the reasoning behind the weight considerations.

Contextual Notes

There is a mention of a diagram that may clarify the setup, which is currently not available to all participants. The discussion reflects varying interpretations of the problem and the principles involved.

Prannoy Mehta
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Homework Statement



A thin uniform circular tube is kept in a vertical plane. Equal volumes of (The liquids subtend a right angle at the centre) two non miscible liquids whose densities are a and b respectively fill half of the tube as shown. (The diagram depicts a>b) In equilibrium the radius passing through the interface makes an angle of 30 degrees with the vertical. The ratio of densities a/b is equal to.

Homework Equations



The basic equations of fluid mechanics given in any introductory course.

The Attempt at a Solution



I tried equating the weights on both the side as it must be necessary condition for the equilibrium.
Doing so I have obtained the answer as 3. The answer given in the text is 3.732. Where is the flaw in my concept.

Thanks for all the help.
 
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How can a tube be kept in a plane?
Is the tube lying on its side or standing on end?
I think we need to see the diagram.
 
I thought it was not required, sorry for the trouble..

upload_2015-10-24_16-8-10.png
 

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The flaw in your logic us using the weights. You ought to be using dp/dz=-ρg, and equating the pressures at the interface.

Chet
 
I still did not understand why that concept of weight is wrong, for equilibrium the weight of the liquid with density a, subtending 60 degrees at the centre, should be equal to that, of the a subtending 30 degrees, and the liquid b which is subtending 90 degrees. I do realize there is a flaw in that concept, just wanted to know what is the flaw..

PS: I got the answer using that method. Thank you :)
 
Prannoy Mehta said:
I still did not understand why that concept of weight is wrong, for equilibrium the weight of the liquid with density a, subtending 60 degrees at the centre, should be equal to that, of the a subtending 30 degrees, and the liquid b which is subtending 90 degrees. I do realize there is a flaw in that concept, just wanted to know what is the flaw..
You are aware that the bottom of the curved tube wall supports part of the weight of each fluid, correct?

Chet
 
Yes, I got it now. I din't think of it that perspective.
 

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