Hydrodynamics physics problem

1. Apr 19, 2013

paras02

1. The problem statement, all variables and given/known data
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

2. Relevant equations

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

3. 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

2. Apr 19, 2013

haruspex

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. Apr 20, 2013

paras02

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. Apr 21, 2013

haruspex

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. Apr 21, 2013

paras02

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. Apr 21, 2013

haruspex

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. Apr 22, 2013

paras02

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. Apr 23, 2013

haruspex

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. Apr 23, 2013

paras02

yes it is right and it is an integer type question

10. Apr 23, 2013

paras02

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. Apr 23, 2013

haruspex

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. Apr 24, 2013

paras02

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. Apr 24, 2013

consciousness

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. Apr 24, 2013

haruspex

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.)