Ratio density of liquid - pressure

[zorro]

Homework Statement

The apparatus shown consists of four glass columns connected by horizontal sections. The height of 2 central columns B and C are 49 cm each. The height of the liquid in A and D is 52.8cm and 51 cm resp (measured from the baseline). The water columns are maintained at temperatures indicated. What is the ratio of density of liquid at 95 C to that at 5 C?

The Attempt at a Solution

Pressure at the bottom of A should be equal to the pressure at the bottom of D as they both are on the same horizontal level.

 

[tiny-tim]

… What is the ratio of density of liquid at 95 C to that at 5 C?
…
Pressure at the bottom of A should be equal to the pressure at the bottom of D as they both are on the same horizontal level.

Yes.

So what is the ratio of the densities?

 

[zorro]

Thanks a lot for looking into this unanswered thread.
ρ1 x g x 52.8 = ρ2 x g x 51
ρ1/ρ2 = 51/52.8 = 0.9659
where ρ1 = density at 95C

Unfortunately, this answer is incorrect :(
Where do you think I am wrong?

 

[tiny-tim]

Hi Abdul!

My previous answer was wrong.

The usual https://www.physicsforums.com/library.php?do=view_item&itemid=115" does not work in this case, because it only applies to steady flow.

ok, I know that everything is static, and "you can't get more steady than static!" …

but in principle it is not steady flow, in the sense that if the fluid moved slightly, the boundaries between the two densities would be in different positions.

(Essentially, Bernoulli's equation is a conservation of energy equation, and takes into account all changes in energy … in steady flow, only the energy at the two ends of any streamline changes, but here the energy at each boundary also changes … so, going across a pair of boundaries with height difference ∆h, there is an energy transfer of (∆ρ)(∆h) which isn't in the usual equation. )

why does Bernoulli's equation work when there's only one boundary, ie if there's only one density on the left and another on the right? i haven't quite worked that out yet

Sooo … there are 3 ways of doing this …

i] Use good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" on each of the four columns … call the upward forces at the bottom of the two left columns T₁, the upward forces at the bottom of the two right columns T₂, and the downward force at the top of the two middle columns T₃.

ii] calculate total gravitational https://www.physicsforums.com/library.php?do=view_item&itemid=269" for a fixed total length for each height difference, and see which one is the lowest!

iii] use Bernoulli's equation anyway, but with a correction term ∑(∆ρ)(∆h) for each of the two pairs of boundaries across column B and column C.

… try anyone of them, and see what you get!

(or, preferably, try all three! )​

I have never seen a problem like this before …

does anyone know of a textbook (preferably free online) that discusses it?

 

[zorro]

but in principle it is not steady flow, in the sense that if the fluid moved slightly, the boundaries between the two densities would be in different positions.

If the fluid is displaced slightly, it executes S.H.M. and then comes to a halt (if there are energy losses) in its original position. I did not get your point here.

(Essentially, Bernoulli's equation is a conservation of energy equation, and takes into account all changes in energy … in steady flow, only the energy at the two ends of any streamline changes, but here the energy at each boundary also changes … so, going across a pair of boundaries with height difference ∆h, there is an energy transfer of (∆ρ)(∆h) which isn't in the usual equation. )

What is (∆ρ)(∆h)? Is it the energy difference due to change in density? What kind of energy is that?

ii] calculate total gravitational https://www.physicsforums.com/library.php?do=view_item&itemid=269" for a fixed total length for each height difference, and see which one is the lowest!

I don't understand what you mean by 'fixed total length'. Please elaborate.

I have never seen a problem like this before …

Well this question is from the prestigious IITs (year 1997). They give such thought provoking questions

 

[tiny-tim]

If the fluid is displaced slightly, it executes S.H.M. and then comes to a halt (if there are energy losses) in its original position. I did not get your point here.

I mean that the boundary (between the densities) moves …

that is not steady flow …

for steady flow (and therefore for Bernoulli's equation to apply), all speeds etc, including the position of the boundary, must be independent of time

What is (∆ρ)(∆h)? Is it the energy difference due to change in density? What kind of energy is that?

∆ρ is the difference between the two densities

∆h is the difference in height between two density boundaries

I don't understand what you mean by 'fixed total length'. Please elaborate.

I can't think of any other way of describing it …

fix a total length, and find the potential energy for that length as a function of the difference in height between the two ends.

 

[zorro]

I mean that the boundary (between the densities) moves …

that is not steady flow …

for steady flow (and therefore for Bernoulli's equation to apply), all speeds etc, including the position of the boundary, must be independent of time

I just know that Bernoulli's equation can be applied to a cross-section of an incompressible, non-viscous, irrotational, steady flowing fluid. I did not read anywhere about the boundaries.
Do you mean that we cannot apply it to the liquid as a whole or a part of it containing one boundary atleast ( in simple words for the liquid in A + B because the densities are different)?

I can't think of any other way of describing it …

fix a total length, and find the potential energy for that length as a function of the difference in height between the two ends.

For the liquid in column A,
Gravitational potential energy = ρ₉₅Agh² where A is the area of cross-section.
What do I do next?

 

[tiny-tim]

Hi Abdul!

I just know that Bernoulli's equation can be applied to a cross-section of an incompressible, non-viscous, irrotational, steady flowing fluid. I did not read anywhere about the boundaries.

"steady" is the problem word …

it means that the flow looks the same at all times …

if you take two photographs of it at different times, everything is the same.

In this case, the boundaries will all be in different places, so the photographs will look different, so it's not "steady".

This of course assumes that the fluid is moving … in a stationary system, one has to imagine what would happen if the fluid did move … in this case, it would not be "steady".

btw, I think your use of "cross-section" is wrong … https://www.physicsforums.com/library.php?do=view_item&itemid=115" applies along a streamline, not across a cross-section.

Do you mean that we cannot apply it to the liquid as a whole or a part of it containing one boundary atleast ( in simple words for the liquid in A + B because the densities are different)?

We can't apply it to the liquid as a whole.

I think we can apply it to a liquid with only one boundary (eg 95° on the left and 5° on the right) … but I haven't worked out why!

But certainly not to two or more boundaries at different heights.

For the liquid in column A,
Gravitational potential energy = ρ₉₅Agh² where A is the area of cross-section.
What do I do next?

You really only need to calculate the difference in potential energy between two configurations.

For a given height difference, calculate the change in PE if that height difference is changed by a small amount h.

(Though you may find method i] easier)

 

[zorro]

btw, I think your use of "cross-section" is wrong … https://www.physicsforums.com/library.php?do=view_item&itemid=115" applies along a streamline, not across a cross-section.

Well if all the conditions of Bernouilli's Principle are satisfied, a cross-section of fluid will consist of many stream-lines and we can apply the principle to them. Across a streamline is more precise though.

(Though you may find method i] easier)

Actually I did not understand how to apply Newton's 2nd law here.
After considering the upward/downward forces as you said, what is the next step?
Do I have to equate these forces with the weight of the liquid column?

Please give some more hints on your iii] method too.

 

[tiny-tim]

Well if all the conditions of Bernouilli's Principle are satisfied, a cross-section of fluid will consist of many stream-lines and we can apply the principle to them. Across a streamline is more precise though.

A cross-section is across the flow, a streamline is along the flow.

Actually I did not understand how to apply Newton's 2nd law here.
After considering the upward/downward forces as you said, what is the next step?
Do I have to equate these forces with the weight of the liquid column?

Yes, use the weights …

it's like those problems with masses joined by strings going over pulleys, where you do F = ma for each mass separately, and eventually eliminate the tensions …

instead of tension, we have pressure.

Please give some more hints on your iii] method too.

hmm … haven't completely worked that out yet.

 

[zorro]

A cross-section is across the flow, a streamline is along the flow.

I was talking about this-

Yes, use the weights …

it's like those problems with masses joined by strings going over pulleys, where you do F = ma for each mass separately, and eventually eliminate the tensions …

instead of tension, we have pressure.

We have pressure, but it doesnot have a direction unlike tension .
Can you just write the equation for column A so that I will get some more idea and proceed to solve?

 

[tiny-tim]

I was talking about this-

A cross-section would usually be drawn very thin, not long like that.

We have pressure, but it doesnot have a direction unlike tension .

Yes it does … once you specify a surface (a cross-section in this case!), the pressure is perpendicular to the surface (in this case, along a streamline!)

Can you just write the equation for column A so that I will get some more idea and proceed to solve?

Nope, you're perfectly capable of doing it (I assume you can do pulley questions?)

 

[zorro]

Yes it does … once you specify a surface (a cross-section in this case!), the pressure is perpendicular to the surface (in this case, along a streamline!)

errm I don't agree with you.
Pressure doesnot have any direction as it is a scalar, even if you consider a cross-section.
It is the force due to pressure that has a direction.

According to wikipedia-
"Pressure is a scalar quantity. It relates the vector surface element (a vector normal to the surface) with the normal force acting on it. The pressure is the scalar proportionality constant that relates the two normal vectors:

[URL]http://upload.wikimedia.org/math/d/2/a/d2a035095f1ff78a27e468cb4abd225e.png[/URL]

The minus sign comes from the fact that the force is considered towards the surface element, while the normal vector points outwards.

It is incorrect (although rather usual) to say "the pressure is directed in such or such direction". The pressure, as a scalar, has no direction. It is the force given by the previous relation the quantity that has a direction. If we change the orientation of the surface element the direction of the normal force changes accordingly, but the pressure remains the same."

Nope, you're perfectly capable of doing it (I assume you can do pulley questions?)

I am getting the answer as 100/101.8 which is exact!
While doing the problem, I figured out a shorter method-
If we consider the horizontal section BC, the gauge pressure should be equal as the liquid is stationary.
i.e. Patm + Pa - Pb = Patm + Pd - Pc

If you solve the equation, you get the same answer!
I think this is more shorter. I would fail in my duty if I don't thank to you - I figured this out only after your help