Determination of density at a depth in a compressible fluid.

[ovais]

Determination of density at a depth in a compressible liquid.

Hi all,

Let us consider a tank containing some compressible liquid with bulk modulus K and having density at its surface s. Consider some part of liquid with volume V, a depth h below the surface. Due to liquid above it there will be some extra pressure dP, acting on the volume V of the liquid considered. This pressure or volumetric stress will cause some volumetric strain(negative) dV in liquid volume V.

The bulk modulus is related with the pressure and volume changes as K= -dPV/dV or
1/K= -dV/dPV. ...(1)

We also know density = mass/volume...
For the sake of analysis we can assume Liquid volume V at the surface(where the liquid is negligibly compressed). If its density is s(as specified already), then we can say:
s=m/V. ...(2)
where m is the mass of liquid with volume V at surface.

To form a similar equation for density at a depth h, let us imagine that the liquid of volume,V is now dipped below a depth h. At this depth the volume is this liquid changes by dV(a negative value) the mass of this liquid will however remains same. This accounts for a change in density say s* being the density at depth h. The equation will then be:

s*= m/(V+dV). ...(3)

from equation (2), m=sV, putting this value in 3 we get

s*= sV/(V+dV). or
V+dV= sV/s*. or
dV= sV/s* -V. or

dV= V(s/s*-1). ...(4)

Putting this value in equation 1,we get

1/K= -V(s/s*-1)/VdP. or

1/K= (1-s/s*)/dP...(5)

Equation (5), will give the density s* at a depth h, provided we know the pressure differenece at depth h.

This is where I am finding difficulty. At one look, one may simply say the pressure at depth h will be hsg. But as a matter of fact the liquid is compressible its density is not s throught. The formula hsg will work only if the density were constant and liquid is incompressible.

I have tried a lot of integration tools but in vain.

Any help will be highily appreciated.

Thanks a bunch.

 

[voko]

You will need a way to describe the compression of the fluid at an arbitrary depth. For example, let $x$ be the depth of a small fluid element from the imaginary surface of the fluid in the absence of gravity. Let then $z(x)$ be the depth of that same element with the gravity turned on.

Then, the pressure at imaginary depth $x$, which corresponds to real depth $z(x)$, is $mgx$, where $m$ is the unstressed density. Because the fluid is in equilibrium, pressure must be balanced by its stress. To compute the stress, consider a slightly bigger fluid element whose top is at $z(x)$ and whose bottom at $z(x + dx) = z(x) + dz$. The stressed height of the element is $dz$, the unstressed height is $dx$ so the stress is $K {dx - dz \over dx} = K (1 - {dz \over dx})$, giving $$

mgx = K (1 - {dz \over dx})

$$

Continue from here.

 

[Chestermiller]

In the equation K= -dPV/dV, V is the local specific volume (i.e., the mass per unit volume). The specific volume is the reciprocal of the density ρ. From the hydrostatic equation, we have:
$$dP=ρgdz=\frac{g}{V}dz$$
If we combine these equations, we obtain:
$$K\frac{dV}{V}=-\frac{g}{V}dz$$
If we integrate this equation, we obtain:
$$V-V_{surface}=-\frac{gz}{K}$$
In terms of densities, this becomes:
$$\frac{1}{s}-\frac{1}{ρ}=\frac{gz}{K}$$
From this, if we solve for ρ, we obtain:
$$ρ=\frac{s}{1-\frac{sgz}{K}}$$
Chet

 

[ovais]

Then, the pressure at imaginary depth $x$, which corresponds to real depth $z(x)$, is $mgx$, where $m$ is the unstressed density. Because the fluid is in equilibrium, pressure must be balanced by its stress. To compute the stress, consider a slightly bigger fluid element whose top is at $z(x)$ and whose bottom at $z(x + dx) = z(x) + dz$. The stressed height of the element is $dz$, the unstressed height is $dx$ so the stress is $K {dx - dz \over dx} = K (1 - {dz \over dx})$, giving $$

mgx = K (1 - {dz \over dx})

$$

Continue from here.

Thanks voko I understand upto the last result you brought. But I still fail to understand how this will solve the problem, see the depth x is imaginary it is just a new variable introduced and simply specifying x as an imaginary depth which corresponds to real depth can be a start but here to find stress or pressure(mgx/K) the story again calls for a value of x.I mean how to continue from here?

I will be thankful to you people if you get me my result.

 

[ovais]

In the equation K= -dPV/dV, V is the local specific volume (i.e., the mass per unit volume). The specific volume is the reciprocal of the density ρ. From the hydrostatic equation, we have:
$$dP=ρgdz=\frac{g}{V}dz$$
If we combine these equations, we obtain:
$$K\frac{dV}{V}=-\frac{g}{V}dz$$
If we integrate this equation, we obtain:
$$V-V_{surface}=-\frac{gz}{K}$$
In terms of densities, this becomes:
$$\frac{1}{s}-\frac{1}{ρ}=\frac{gz}{K}$$
From this, if we solve for ρ, we obtain:
$$ρ=\frac{s}{1-\frac{sgz}{K}}$$
Chet

Chestermiller, first of all thankyou. This is really a good easier method, But hold a minute, you are saying specific volume is the reciprocal of density and wrote 1/V for density

But reciprocal of density should beV/m. And It density in that equation should be replaced by m/V not simply by 1/V as you did. I think you miss to put m over here. Now if I put m in the same equation, and one more thing what actually is m? Is it the value of mass of liquid stands on height z above occuping volume V? The main question how can I quantify it?


Further an important falasy I can see in your attempt is that in LHS,at KdV/V, V and dV are actually the volume of liquid under stress at a depth z below where as in RHS, at gdz/V, V is the volume responsible for pressure that is it is the volume of liquid stands from depth h to the the liquid surface on the stressed liquid. While you seem to mixed both,as the same thing and canceled them with each other.

 

[voko]

Thanks voko I understand upto the last result you brought. But I still fail to understand how this will solve the problem, see the depth x is imaginary it is just a new variable introduced and simply specifying x as an imaginary depth which corresponds to real depth can be a start but here to find stress or pressure(mgx/K) the story again calls for a value of x.I mean how to continue from here?

The good thing about the imaginary depth is that it is known in advance and does not depend on the details of the compressed state. It changes in the interval $[0, p]$ where $p$ is the maximum depth. Note that the "imaginary" max depth and the "real" max depth are the same depth, which gives $z(p) = p$. This, together with the final equation of my previous post, fully defines $z(x)$ and knowing $z(x)$ you can find the density at $x$ (or at $z$).

 

[ovais]

The good thing about the imaginary depth is that it is known in advance and does not depend on the details of the compressed state. It changes in the interval $[0, p]$ where $p$ is the maximum depth. Note that the "imaginary" max depth and the "real" max depth are the same depth, which gives $z(p) = p$. This, together with the final equation of my previous post, fully defines $z(x)$ and knowing $z(x)$ you can find the density at $x$ (or at $z$).

volvo, how can we say that imaginary and real,depths are equal, when we know the fact the liquid even above the "liquid considered at depth z?
With reference to your first reply let say I have to find the density at a depth z (from the real surface of compressed liquid and here strictly z and x in itself can't be equal or same) What value x ,in mgx= K(1-dz/dx) do I need to take? After knowing nowing this, also tell me how do I actually have to workout on the relation mgx=K(1-dz/dx).

 

[voko]

volvo

It is voko :)

how can we say that imaginary and real,depths are equal, when we know the fact the liquid even above the "liquid considered at depth z?

I am not entirely sure what you mean here. My point of reference is the location of the surface of the uncompressed fluid. I label that point with $0$. The values of both $x$ and $z$ are measured from this point down. Going down from that point, we reach the bottom at $p$. The bottom obviously stays at $p$ no matter whether the fluid is compressed or not, so at the bottom we must have $z(p) = p$. At the surface the situation is different: $z(0)$, which gives the location of the surface of the compressed fluid, is always below the surface of the uncompressed fluid, so $z(0) > 0$. $p - z(0)$ then gives the entire height of the column of the compressed fluid as seen from the bottom. I hope that clarifies things.

 

[ovais]

It is voko :)

Oh sorry :)

The bottom obviously stays at $p$ no matter whether the fluid is compressed or not, so at the bottom we must have $z(p) = p$.

What does it mean?

the compressed fluid, is always below the surface of the uncompressed fluid, so $z(0) > 0$. $p - z(0)$ then gives the entire height of the column of the compressed fluid as seen from the bottom. I hope that clarifies thin]

This fact that the compressed fluid is always below the uncompresd fluid(if liquid above were uncompressed) is what posing difficulty to me. As I don't know how much more depth should I take to take an account of it.For example if I have find density at a depth of 20m below the surface specified by 0, where liquid is uncompressed. As you point out the density keeps on increasing till z(p), p being bottom. Here z(real depth) is 20m but if the liquid we not compressed this would( you specified it by x) definitely less than 20m. This force me to think,that actually z and x are not same.

I am,sorry for my weak Mind, due to which I am getting things very slowly. I am thankful to you for your cooperation. :)

 

[voko]

It just crossed my mind that the bulk modulus is defined with respect to the change in volume, not to a linear dimension as I used in my derivation, which invalidates it. So I just wasted your time and I am sorry about that. Please follow Chet's method.

 

[ovais]

It just crossed my mind that the bulk modulus is defined with respect to the change in volume, not to a linear dimension as I used in my derivation, which invalidates it. So I just wasted your time and I am sorry about that. Please follow Chet's method.

Ok, its Ok. In your method I thought of multiplying by area(which is constant for a tank,of uniform cross-sectional area) so that your linear dimension get itself convert to volume. can you notice the mistake in chet's attempt. He writes specific Volume(reciprocal of density) to be 1/V instead of V/m.

Also in derivation he mixed volume of compresed liquid under stress and volume of liquid stands above it and take both of them as V although these two are different. As I pointed out in my then next reply to him.
Do you have any idea how to proceed even with his method?

 

[voko]

Ok, its Ok. In your method I thought of multiplying by area(which is constant for a tank,of uniform cross-sectional area) so that your linear dimension get itself convert to volume.

Unfortunately, my method would still be wrong even in this case. The problem with it is that it assumes that the fluid is compressed solely in the vertical direction, and so the pressure at depth $z$ is given by $mgx$, with the $x$ corresponding to the $z$. That is not correct for a fluid that is compressed uniformly, because there is no one-to-one dependence between $z$ and $x$.

I will look more closely at Chet's method.

 

[ovais]

Unfortunately, my method would still be wrong even in this case. The problem with it is that it assumes that the fluid is compressed solely in the vertical direction

I think this is not a problem: in hydrostatics though pressure changes in depth are given in terms of vertical height(h) but this remains compatable with the fact the pressure at any point within the fluid acts in(from) all directions.

Don't know is there any solution for such problem. :(

 

[voko]

To make Chet's method more explicit, let's label the "specific volume" with $v$, and use $V$ to mean just "the volume". The specific volume is defined as the ratio of volume to mass, so for any volume $V$ with mass $M$, $v = \frac V M$. Density is defined as $\rho = \frac M V$, so $\rho = \frac 1 v$. In the equation $K = - dP \frac V {dV}$, divide the numerator and the denominator with $M$, obtaining $K = -dP \frac v {dv}$. So now everything is expressed in terms of the specific volume, and the rest of Chet's derivation follows.

 

[ovais]

To make Chet's method more explicit, let's label the "specific volume" with $v$, and use $V$ to mean just "the volume". The specific volume is defined as the ratio of volume to mass, so for any volume $V$ with mass $M$, $v = \frac V M$. Density is defined as $\rho = \frac M V$, so $\rho = \frac 1 v$. In the equation $K = - dP \frac V {dV}$, divide the numerator and the denominator with $M$, obtaining $K = -dP \frac v {dv}$. So now everything is expressed in terms of the specific volume, and the rest of Chet's derivation follows.

In this attempt you are making the same mistake as chet did but in a different style.

See we assume a volume V deep h, belwo the water surface. We can further assume the mass of this volume as M. So that the specific volume be V/M. This is ok. But in expression for pressure dP= dgdz, the density d which is varying, is for the entire liquid above that of volume V below we considered. And for replacing density d with specific volume in this case(v=V/M) the volume of the liquid above may not be V also the mass M is specially the mass of liquid below at depth under stress and it may not be the case that the mass above also be M and volume V. In fact the specific volume v liquid under depth is assumed to be constant, while we still fail to make expression for specfic volume for replacement of density, in pressure dP expression unless we create an new variables M' and V' as mass and volume of liquid above the liquid volume V considered. Or we put v' as a variable other than v(constant specific volume) for replacement of density(the variable).

 

[voko]

But in expression for pressure dP= dgdz, the density d which is varying, is for the entire liquid above that of volume V below we considered.

No, that is not true. First of all, $d$ here is not a density. It is part of the symbols $dP$ and $dz$, which mean "a tiny change in pressure" and "a tiny change in depth", respectively (more accurately the "tiny change" thing is known as the "differential" in calculus). You cannot consider the $d$ in these symbols separately from $P$ and $z$.

So we are relating a tiny change in pressure over a tiny layer of fluid with the tiny change of depth over said layer; we do not care (in this equation) about the entire column above or below the layer.

 

[ovais]

Ok based on the introduction of specific volume v=V/M and chet's equations, how you will help me go on.

K= -dPV/dV, V is the local specific volume (i.e., the mass per unit volume). The specific volume is the reciprocal of the density ρ. From the hydrostatic equation, we have:
$$dP=ρgdz=\frac{g}{V}dz$$

Now that you have form something like v, now how you will like to combine chet's equations proceed?

 

[voko]

Just replace $V$ with $v$ in Chet's equations and you are done. The final equation that Chet obtained gives you the density as you wanted ($s$ is the density at the surface, or the density of the uncompressed fluid).

 

[ovais]

The two v's are not equal! One v is the v of the liquid under depth h and has a constant value while the other v is of liquid standa above it and it is a variable. You are losely looking at them and thinking they are same but they are not.

The replacement of roh(d) by v is not as simple. I think you have to look closely.

 

[voko]

The two v's are not equal!

Chet used the capital $V$ to mean "specific volume". I used $v$ to denote the specific volume because it seemed to me you were confused by the similarity of $V$ with "volume" (without the "specific" qualifier). So yes, they mean one and the same thing if you read carefully what Chet and I wrote.

One v is the v of the liquid under depth h and has a constant value while the other v is of liquid standa above it and it is a variable.

That was not said by either Chet or me. $v$ in my notation and $V$ in Chet's notation is the specific volume, a function of the depth $z$, which is definitely not a constant value.

 

[ovais]

$$dP=ρgdz=\frac{g}{V}dz$$

isn't ρ variable here?? Or a constant? For me it is a variable in this differential equation. And if ρ is a vraiable so should its specific volume v. Isn't?

Now tell about this

$$K\frac{dV}{V}=-\frac{g}{V}dz$$

is V variable or constant? For me I believe here V was the volume(before compression) of liquid at depth h and it belongs to that part of liquid only. And dV is the small change in its volume due to pressure. And if V is constant so should its specific volume v. So the v of this is constant while the v of that is constant and we can't call them with same v. frankly speaking it is all confusing :)

 

[ovais]

Forget about (capital) V , I am saying even the two,specific volumes(small v) are not the same both of which you represent by small v.

 

[voko]

$$dP=ρgdz=\frac{g}{V}dz$$

isn't ρ variable here?? Or a constant? For me it is a variable in this differential equation. And if ρ is a vraiable so should its specific volume v. Isn't?

Everything except $g$ is a variable in that equation. More accurately it could be said that everything except $g$ is a function of $z$. Of course, when you make $z = \mathrm{const}$ all that becomes constant, so the distinction is not absolute.

Now tell about this

$$K\frac{dV}{V}=-\frac{g}{V}dz$$

Same story. $V(z)$ is the specific volume at depth $z$. $dV$, which in its full form is $dV(z)$, is the differential of $V(z)$, which could be equivalently represented as $V'(z) dz$, where $V'(z)$ is the derivative of $V(z)$. Whether all these are "variables" or "constants" depends on your point of view, but the important thing is that they are all evaluated at one and the same value of $z$, and whatever happens at other values of $z$ does not enter the equation.

 

[ovais]

Oh I understand these are just specific volume both at any depth z and it can be used in both as such.

Thanks got it:)

 

[ovais]

But in chet's method I think there is a problem at the end part of derivation for replacing specific volume v1 and v2 with densities. See


$$V-V_{surface}=-\frac{gz}{K}$$
In terms of densities, this becomes:
$$\frac{1}{s}-\frac{1}{ρ}=\frac{gz}{K}$$

remember here we will get same result but to convert to density we have to put mass m somehow? Like this
$$\frac{m}{s}-\frac{m}{ρ}=\frac{gz}{K}$$

now I want to know what mass will then I need to put.

 

[voko]

As explained above, the specific volume is simply the reciprocal of density. So $$V = \frac 1 \rho.$$

 

[ovais]

Ahhaa correct :)

 

[ovais]

If the specific volumes variable v used two equations same thing, I have an interesting question.

suppose I have a drop of liquid1 (having density d1 when uncompressed) bulk modulus K1 with volume V, immersed in another liquid of density d2 (bulk modulus K2) at a depth z. What is the density say d1* of the drop(liquid1) under this condition?

How can one go on putting variables in this case the specific volume(variables) for liquid1 and liquid2 in this case will definitely not the same.

My Mind is saying that in previous attempt we did take bulk modulus only for stressed volume at the depth while completely ignore the bulk modulus(or compression in column of liquid above that). Because if we compare the final result for density at the depth it is coming same as one get if he/she assume pressure at depth z to be zdg, d being the density of fluid(in the column) in uncompressed state.

 

[voko]

This is becoming (much) more complex. So far we have been looking at an equilibrium situation. When would a drop of liquid immersed in other liquid be in equilibrium?

 

[ovais]

This is becoming (much) more complex. So far we have been looking at an equilibrium situation. When would a drop of liquid immersed in other liquid be in equilibrium?

Ok let me make it easy. Suppose I have a balloon filled with 1000cm3 water at depth of say 20m in a water. If I want to observe change in its density(from 1g/cm3 at surface) I can use equation K= -dPV/dV
where dP is the pressure at depth of 20m and V is the volume of water in balloon(1000cm3) dV is the change in volume of WATER IN BALLOON.

so I need pressure at depth of 20m.
I know we can change the above equation in terms of specific volumes, that would be perfect. But I would like to clarify that do we have to use equation
K=-dPV/dV
for whole water or only to the water in balloon. In this case?
is

 

[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.

 

[ovais]

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.

 

[ovais]

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.