Pressure under water considering bulk modulus

Question

I want to calculate the under water pressure, taking into account the compression of water. I derived a formula, but apparently the function has a vertical asymptote, meaning that the pressure approaches infinity at a certain depth under water. What is my mistake?

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Let's say the atmospheric pressure is $p_o$, and $h$ is the depth of water, and $\beta$ is the bulk modulus. The increase in pressure in an infinitesimal increase in depth($dh$) can be measured by $\rho g.dh$. So:

$dp = \rho g.dh$
$\beta = -\dfrac{dp}{dV}V => -\dfrac{dV}{V}=\dfrac{dp}{\beta}$
$\therefore V(p)=V_0 \times e^{\dfrac{-p}{\beta}}$
$\rho = \dfrac{mass}{volume} => \rho = \rho_0 \times e^{\dfrac{p}{\beta}}$
$\rho_0$ is the density of water at $h=0$
$dp=\rho_0 \times e^{\dfrac{p}{\beta}}\times g .dh$
Now I integrated the both sides and solved for $p$ as a function of $h$. The result was:
$p = -\beta \times ln(e^{\dfrac{-p_0}{\beta}}-\dfrac{\rho_0 gh}{\beta})$

But this function has a vertical asymptote, which means pressure approaches infinity, likely a wrong result.

Thanks!

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The only explanation I could come up with is that the bulk modulus is not a constant number.

Answer 1

The value of $\beta$ for water is 2.2 GPa, and the greatest depth of the ocean is only 10000 m. So the value of the 2nd term in parenthesis in your equation is less than 0.05, even at the very bottom of the greatest ocean depth. This compares with a value of essentially 1.0 for the first term in parenthesis. So the limiting depth that you refer to is not even close to being approached in the real world. In essence, the compressibility of water can be neglected in this instance on a practical basis.

Answer 2

Your mathematical result is clearly incorrect. The pressure at the bottom is dependent only the mass of water in the column. You cannot get an infinite pressure from any finite amount of water. It would seem that the problem lies with the use of the depth of the water. Consider this way of experimentally testing your equation:

Take a cylinder, with a convenient $1$ $m^2$ cross-section, and dump a little water into it.

Measure the depth, and the pressure at the bottom.

Now dump in a standard $1$ $m^3$ of water. Observe very closely what happens.

As a first approximation, the water gets $1 m$ deeper.

But...

Each of the cubic metres below the old surface is now subject to a greater pressure because of the added pressure from the increased mass of water above it.

So each cubic metre of water in the column shrinks a little. Not much, but a little. So the level, for a split second $1 m$ higher, drops slightly

So just keep on adding a cubic metres of water, and observing the pressure. Also keep an eye on the water level!

Eventually the column will be so high, with so many cubic metres of water in it, that the additional pressure throughout the column, when one cubic metre water is added, will cause more than one cubic metre of total shrinkage throughout the column. Voila! The depth decreases!. And continues as you add more and more water.

The reason that some depths lead to impossible pressures is that some depths are impossible.