Hydrodynamics physics problem

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Homework Statement


A beaker contains a liquid of density ρ up to a height h (from the bottom), their is another layer of liquid of density 2ρ above it.(this is arrangement initially) also up to a height h. A hole is made in the upper part as well as in lower part such that the separation between them is h. Speed of the water flowing through lower part is v1 and that from upper part is v2. Find v12/v22


Homework Equations



Bernoulli theorem and speed of efflux
i.e p + 1/2ρv2 + ρgh = constant
and v2 = 2gh respectively

The Attempt at a Solution


don't know what's going to happen as liquid of high density is above the liquid having lower one
 

Answers and Replies

  • #2
haruspex
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Could you describe the set up in more detail, pls? Where exactly are these holes - in the side of the beaker? What stops the fluids intermixing within the beaker? Is there some barrier?
 
  • #3
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No there is no such barrier. This is the initial setup maintained somehow and holes can be any where in upper part and lower part such that separation between them is h.
 
  • #4
haruspex
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OK, I think I get the picture, but I'm going to assume that by some magic the two liquids do not mix.
If there were just one liquid and the hole were h below the surface, what would the speed be?
 
  • #5
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I think it will be square root of 2gh for upper hole and zero for the lower one. But practically it is impossible to make the hole at zero height from the bottom ,so we can assume that holes are somewhere between the upper and lower part respectively.
 
  • #6
haruspex
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Not sure what you mean by "zero for the lower one" here since I was posing a scenario with only one hole. no matter, we can indeed take the velocity as being proportional to √gh. Now with the two liquids, suppose the depths of the holes below the top surface are x and x+h. What is the speed out of the upper hole?
The lower hole is trickier. What is the pressure in the beaker at the lower hole?
 
  • #7
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i think the pressure at the lower hole is (2ρ)gh + ρgx
Considering a point 'M' ,h height below the surface where the two layers meet the pressure is density*g*h. Now the height of the second hole from the bottom is { 2h - (h+x) } i.e h-x also the height of lower part is h therefore its height from point 'M' is { h-(h-x) } so pressure at lower hole is ( pressure at 'M' + ρgx )
 
  • #8
haruspex
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I think there must be something in the set-up I still have wrong. The way I read it, we're not told how far the top hole is from the top surface, so it could be zero. That would make v1=0, so the ratio depends on that distance.
Have you quoted the exact wording? Maybe each hole is supposed to be at the bottom of the section it's in?
 
  • #9
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yes it is right and it is an integer type question
 
  • #10
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I think it is a general question therefore the ratio will remain same for any height. So we can make it specific.
What will be your answer if the hole is h/2 height below the surface ? I have chosen h/2 bcoz i think it will make the whole setup symmetrical.
 
  • #11
haruspex
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I think it is a general question therefore the ratio will remain same for any height.
But it clearly does not. If each hole is at the top of its allowed range then there is almost no flow from the top hole. If you disagree, please post a diagram of how you interpret the question.
 
  • #12
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I think it is sensible to conclude bcoz when we increase the height of the upper hole (taking reference of the upper surface) velocity of liquid from both the holes will increase simultaneously and the ratio will remain constant.
You are also right as we have to make certain assumption and neglect the end effects.
Alter the question if it seems right to you (i.e consider h/2 distance) of the hole from the top otherwise we have no way left. Pls try to reply with the answer.
 
  • #13
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There are obviously errors in this question. Not only must the position of atleast one of the holes must be specified, how is it possible that when two immiscible liquids are mixed the lighter liquid is actually below the heavier one ? Perhaps it is best to leave this question.
 
  • #14
haruspex
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There are obviously errors in this question. Not only must the position of atleast one of the holes must be specified, how is it possible that when two immiscible liquids are mixed the lighter liquid is actually below the heavier one ? Perhaps it is best to leave this question.
Arguably there's no difficulty with the denser liquid being on top. The OP does say that this is just an initial arrangement. That the arrangement is not stable should not affect the speeds in the short term. But I agree that either there's some information missing or the set-up is not as it seems. (The wording is a bit strange, like it's a translation from another language.)
 

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