# 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.