Homework Help: Density of water underwater

1. Sep 26, 2009

LBloom

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

Estimate the density of the water 5.3 km deep in the sea. (bulk modulus for water is B=2.0 *10^9
2. Relevant equations

$$\Delta$$V/V=($$-1/B$$)*$$\Delta$$P

dP/dY=-$$\rho$$g

3. The attempt at a solution

I tried to use the equation:
P=Po*e^(-$$\rho$$o/Po)*gh
But i kept getting ridiculously large answers that i dont think could possibly be true.

I said that the initial pressure was 1atm or 1.013*10^5 Pa and that initial density was 1*10^3(kg/m^3).

I know that once i get the pressure under water, I can find the change in volume divided by volume (delta V over V) and from there figure out how what the density of a theoretical block of water would be if submerged 5300 meters. Its just finding out the pressure that trips me up. Am i using the wrong constants or must i derive a different equation for pressure as a function of height?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 26, 2009

Delphi51

The pressure is the force of gravity on a column of water divided by the area of the base of the column. If you consider a column with area 1 m^2 it should be easy to get an expression for the pressure at any depth. As a check, you get roughly an extra atmosphere (about 100 kPa) for every 10 meters of depth.

3. Sep 26, 2009

LBloom

Thats assuming that density is constant. The question is asking for the density under this depth of water

4. Sep 26, 2009

Delphi51

Oh, that makes it delightfully complicated! A little out of my depth, but my intuition says you must work with a bit of depth dh and get an expression for the dP across that distance that incorporates the change in density due to the dP.

5. Sep 26, 2009

LBloom

I think you just set me on the right track :), not sure though

Last edited: Sep 26, 2009
6. Sep 26, 2009

Delphi51

I've been playing with it a bit. I'll write down my doodling in case it is useful. I'm working with the column of water with area 1 m^2.
The bit of height is dh, with mass dm = p*dh (I can't write a rho here). dP = g*dm = gp*dh (1)

Somehow we have to get that B modulus in there. I see it is defined in Wikipedia as B = -V*dP/dV. Got to get rid of the V, I think, so using p = m/V or V = m/p and differentiating I get
dV = (p*dm - m*dp)/p^2. Subbing that into the B definition I get
dP = (Bm*dp - Bp*dm)/mp. (2)

What happens if we sub (2) into (1)? Perhaps it will make a differential equation that can be solved for the density.

7. Sep 26, 2009

LBloom

it might work, but how'd you get dP=g*dm? wouldn't that be in units of force?

8. Sep 26, 2009

Delphi51

(2) is not making sense because of the m, which comes from the V in the B definition. Maybe the whole approach is a failure. There needs to be something in the diagram to make sense of the dp and I rather think there is going to be a second derivative in the equation, if we ever find it. I'm going to sleep on it.

9. Sep 26, 2009

Delphi51

Well, it is over the 1 m^2 area . . . and the next line, dP = gp*dh is one of the formulas you listed in the question.

10. Sep 26, 2009

LBloom

ah, it was a typo (dP=g*dm instead of dP=gp*dh)
Second thing: dm=pdV, not dm=pdh since p is in units of kg/m^3

I've playing around with it also, but i cant seem to get it to work. Theres an equation for pressure as a function of height i wrote above that i might work if i edit it and sub in some of the equations from the bulk modulus. I'll probably finish this prob tommoro

11. Sep 26, 2009

Delphi51

Hmm, we may be out of our "depth" here. Take a look at the graph of density vs pressure here:
http://oceanworld.tamu.edu/resources/ocng_textbook/chapter06/chapter06_05.htm

Upon rereading your question, I see it says "estimate". That suggests we should not be doing all this work! Maybe take that B definition and replace dP with the pressure you would expect at 5 km depth without considering density. Apply to a cubic meter of water. First order approximation? Check the answer on the graph?

12. Sep 27, 2009

LBloom

omg. I tried that in the beginning, but the computer said i was wrong. I just realized, after i finished calculating the change in volume i used the wrong density. I gave an arbitrary value for V (1 cubic meter) and said the original mass must have been 1000 kg (which i divided by the new volume to get the new density) However, the question is asking about the ocean and the original density of seawater is 1025 kg/m^3, not 1000. I changed that and now the computer says i was right.

thank you! hope you didnt mind me wasting your time with differentials :)

13. Sep 27, 2009

Delphi51

Oh, I enjoyed the whole thing! But I have lots of time.

14. Dec 8, 2010

potto

we have the solution

Recently this problem was solved. The governing equation leads to an integral equation which has an exact analytical solution.
see here http://www.potto.org/downloadsFM.php in static chapter and some in the introduction.

good luck!