Determination of density at a depth in a compressible fluid.

[voko]

If there is water inside and water outside, and the balloon's surface provides no extra pressure, then there is no difference between the water inside and water outside, so the same equation works throughout.

 

[Ahsan Khan]

If someone is interested to find the pressure at a depth z below a compressible liquid with density(of uncompressed state) at surface as d. Bulk is known as K. How then one go on to find pressure at a depth z?

 

[voko]

The original equation involving pressure was $dP = \rho g dz$, so $$ P(z) - P(0) = g \int\limits_0^z \rho(z) dz .$$ You know $\rho(z)$, so you just need to integrate.

 

[Ahsan Khan]

You know $\rho(z)$, so you just need to integrate.

The problem is that we don't know $\rho(z)$. What we only know is $\rho(z)$ is a function of z but don't know how can we write it in terms of z, so that we can put it under integral, as we don't the caluculus of several variables. Can you do me a favour by putting $\rho(z)$ in suitable manner that allows us to integrate the above equation for determining pressure.

 

[voko]

The problem is that we don't know $\rho(z)$. What we only know is $\rho(z)$ is a function of z but don't know how can we write it in terms of z

What are you talking about? Chet gave you the expression for $\rho$ in terms of $z$. Use it.

 

[Ahsan Khan]

So we have to go back to that? I was thinking of applying the equation K= -dPV/V locally to the compressed portion of the liquid at depth z, where if we are able to find pressure. Once by finding presaure some other method(without using density equation) we can use K= -dPV/V locally to get density of compressed liquid under consideration at depth z.

 

[voko]

I do not understand what you are trying to do here. In #3, Chet derived the expression for density in terms of depth. In #33, I gave you the integral that gives you pressure by using that expression. Substitute, integrate, done. No need to make things more complex than they are.

 

[Ahsan Khan]

I did the integration and with due care I get the result for pressure change as below:

P(z) - P(0) = Kln(K/(K- sgz))... (ln= natural log)
s being the density of uncompressed liquid.

This is pressure difference from the liquid at depth z and pressure at top surface. Right?

And chet's result is as follows:

ρ= s/(1- sgz/K)



Now I want your attention to a problem in my textbook.

The problem (solved) is as follows:

The density of ocean water at the surface is 1030kg/m3. What is the density at a depth where pressure is 80.0 atm. Given that Bulk modulus = 2.18*10^9Pa and 1 atm= 1.013*10^5Pa, K= 2.18*10^9Pa.

solution.
Given s= 1030kg/m3, pressure at the depth where density is to be found)= P=80.0*1.013*10^5 =81.04*10^5Pa, ρ=?

ρ= s/(1-P/K)= 1034 kg/m3( on putting values). Same formula as of chet's using P(pressure at depth which is already given in problem to equal to 80.0 atm).

Now my question is chet use the formula as:

ρ= s/(1- sgz/K)

where as my text is using it like this:


ρ= s/(1-P/K). Here they use P(instead of sgz)which is the pressure at the depth where density is to be found.

It is imparitive from the above that P= sgz and it is basically the pressure at the depth z where density is to be found.

But if we workout for pressure at depth z then we are getting an expression other than sgz. On integrating we are getting it in terms of log(ln).
P(z) - P(0) = Kln(K/(K- sgz)) which is not sgz? That we see in chet's formula. I mean how will you explain this.

 

[Ahsan Khan]

They actually ain text problem also derived this formula in the solution for the present situation.

They did it like this:

Let V and V' be the volume of a certain mass of ocean at the surface and at the depth respectively and s and ρ the coresponding densities of ocean water. Then

V= M/s and V'= M/ρ the decrease in volume at the depth is
-v= V-V'= M( 1/s - 1/ρ)

therefore the volume strain is
-v/V= M/V( 1/s - 1/ρ) = s( 1/s - 1/ρ) = 1- s/ρ)...(1)

by definition, the bulk,modulus of ocean water is

K= Pv/V
or
-v/V= P/K. ...(2)

From equation (1) and (2) we have

1- s/ρ = P/K or

ρ= s/(1- P/K)

same as chet's result but here it is remarked that the numerator of K is actually the pressure at depth z.

But when I did calculations using chet's density formula as above I get something in logrithims.

 

[voko]

The problem with the book's derivation is that it uses finite differences where differentials are supposed to be used. Which may actually be OK for water, given its very high bulk modulus. I suggest that you obtain numeric results using the book's and Chet's formulas, and compare. I suspect they should be very close to each other.

 

[Ahsan Khan]

Ok I will do it and will let you know tomorrow(it is about 11:30 pm) morning, as I have some other problems for today :)

 

[Ahsan Khan]

First of all just one correction(I edited the already correct expression, while writing solution of book's methd, wrong K=-P/v/V (which was wrongly edited to. K=-Pv/V rest was all ok)


Ok I now have done a comparison using book's and chet's formula.

the final result of density as per book is coming as 1034kg/m3(as calculated in previous posts)


For using chet's formula,everything is same except that in the expression ρ = s/(1-sgz/K) the numerator of K doesn't contain the pressure at depth z rather it is sgz.

as here we do not know depth z, but we do know that they(book) use pressure at depth z as a numerator of K in formula. And the pressure(actually it is difference of pressure form that at top, as indicated by other examples of the book) is given as 81.04*10^5Pa.

And we also have derived an expression for pressure at depth z as P(z) -P(0)= Kln(K/(K-sgz)


Now if we equate Kln(K/(K-sgz))= 81.04*10^4

and solve it we are getting

sgz= 8088956Pa.

and we know use this value of sgz in chets formula, the value of density I am getting is 1033.836kg/m3.As opposed to book's method where the answer is 1034kg/m3. The values as you suspect are very close to each other.!

Now can you explain what actually was mathematical or physica problem in book's method which can explain why the value is slightly more.

I have this question to understand, does the value by finitr difference method ALWAYS come larger? I want this answer with reason.

Thanks a lot for continues support:)

 

[voko]

The problem is, as I said, they used finite differences, while the bulk modulus is defined in terms of infinitesimal differences. See, for example: http://en.wikipedia.org/wiki/Bulk_modulus.

From the mathematical point of you, the exact result for the dependence of pressure on depth is $$P - P_0 = - K \ln \left( 1 - {\rho_0 g \over K} z \right) $$ The first derivative of the expression on the right is $$ \rho_0 g K \over K - \rho_0 g z $$ and the second derivative is $$ \rho_0^2 g^2 K \over (K - \rho_0 g z)^2 $$ thus the approximation of the exact result is $$ P - P_0 = \rho_0 g z + {(\rho_0 g)^2 \over 2K}z^2 + ... $$ The book essentially only used the first term of this approximation. The first term, evaluated at the depth of 1000 meters, is about 10 MPa, while the second term there is only about 24 kPa, three orders of magnitude less. Further terms will be still smaller.

It can be said, however, that in general the pressure calculated with the exact formula is always a little greater than the pressure according to the approximate formula.

 

[Ahsan Khan]

the exact result for the dependence of pressure on depth is $$P - P_0 = - K \ln \left( 1 - {\rho_0 g \over K} z \right) $$

This is exactly what I obtained they just inverse the quantity in log by putting a minus sign before log.

The first derivative of the expression on the right is $$ \rho_0 g K \over K - \rho_0 g z $$ and the second derivative is $$ \rho_0^2 g^2 K \over (K - \rho_0 g z)^2 $$ thus the approximation of the exact result is $$ P - P_0 = \rho_0 g z + {(\rho_0 g)^2 \over 2K}z^2 + ... $$ The book essentially only used the first term of this approximation. The first term, evaluated at the depth of 1000 meters, is about 10 MPa, while the second term there is only about 24 kPa, three orders of magnitude less. Further terms will be still smaller.

It can be said, however, that in general the pressure calculated with the exact formula is always a little greater than the pressure according to the approximate formula.

very well explained thankyou, wikipedia also is at same instanse.
:)

 

[Ahsan Khan]

Just to remove any trace of doubt, I want your response for a question that comes to my Mind:

we noticed in the problem in book where the pressure at a depth is 81.04*10^5 Pa, and we had to find the density at this depth, when the density at surface(density of uncompressed liquid) is given 1030kg/m3. And by chet's formula it was coming 1033.8. Kg/m3(approx) and 1034 kg/m3 in book's formula.

if I have a problem like say I have a very small size balloon filled with water at 1030 kg/m3 density and a pressure of 81.04*10^5 Pa is acting on it. There is water column no height only pressure is acting and that too on a very small element(samll size balloon).

Can we work for change in volume or density by appling bulk modulus definition locally to the small balloon? Also will in that case do we necessarly take V as variable(means can we take V as an initial volume befor application of excess pressure start appling bulk,modulus locally as here dP is just fixed not varying) And most importantly does even in the balloon(upon which a constant know pressure is acting) the density is varying within the volume of balloon?

 

[Chestermiller]

Let's take my equation ρ=\frac{s}{1-\frac{sgz}{K}}, which is more precise than the book result, and expand my equation in a Taylor series in s. We get:


ρ=s(1+\frac{sgz}{K}...)

If we linearize the summation with respect to s, we get:

ρ=s(1+\frac{sgz}{K})

I presume this is the result in your book, which is only approximate.

Chet

 

[Ahsan Khan]

ρ=s(1+\frac{sgz}{K})

I presume this is the result in your book, which is only approximate.

Chet

No the result is ρ= s/(1-P/K), where P is the pressure at depth where density is to be found.

and in your equation the numerator of K is sgz, which actually is not the pressure at depth z.

And shown by voko the actual pressure -Kln1- sgz/K)) is somewhat greater than sgz and the book uses the approximate value sgz, which he shown using Taylor or Macalurin theorem.

This is all clear. This time I am talking about density change when a definte pressure(externally) is acting on water in a balloon. See my that post.