Finding Speed of Efflux for Two Liquids of Different Densities

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Homework Help Overview

The problem involves determining the speed of efflux for two immiscible liquids with different densities contained in a vertical container. The setup includes a hole in the container and requires consideration of the pressures and densities involved in the efflux process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the pressure difference at the hole and how it relates to the densities of the liquids. There is an exploration of how to account for the changing density of the liquid as it flows out and the implications of the height variable 'h' in the calculations.

Discussion Status

The discussion is ongoing, with participants questioning the assumptions made regarding the density of the liquid flowing out and the initial conditions of the problem. Some guidance has been offered regarding the expressions for pressure and velocity, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants note that the problem involves two different densities and that the height 'h' is variable, which complicates the analysis. There is an acknowledgment that the density of the liquid flowing out may change over time, which has not been fully addressed in the current discussion.

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Homework Statement


The container of uniform cross section area A holds two immiscible non-viscous incompressible liquids of densities d and 2d each of height H/2. A tiny hole of area S<<A is punched on the vertical side of the container at height h(<H/2)

Find the speed of efflux.

Homework Equations





The Attempt at a Solution



I know how to find it when there is only one liquid present.
What is the approach in this case?
 

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The pressure difference at the hole is
P - Po = g[H/2*d + (H/2-h)*2d ]

The density of the liquid flowing out is 2d. Hence the velocity of the liquid is

v = sqrt[2(P - Po)/2d]

Substitute the values and find v.
 
Last edited:
rl.bhat said:
The pressure difference at the hole is
P - Po = g(H/2*d + H/4*2d) = gHd

how did you get H/4*2d? the height 'h' is a variable.

rl.bhat said:
The density of the liquid flowing out is 2d

After some time density of liquid flowing out will be d. But we are not taking it into account. Why is it so?
 
Abdul Quadeer;2906559[B said:
]how did you get H/4*2d?[/B] the height 'h' is a variable.

It should be (H/2 - h)*2d

After some time density of liquid flowing out will be d. But we are not taking it into account. Why is it so?

From the expression we get only the initial velocity. As the level of the liquid decreases, the velocity will also change. When the liquid with density d flows out, again velocity will change.
 
Thanks a lot, Sir.
 

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