Physics problem: ocean compressibility

This problem is not complicated, but it is particularly interesting to me as it is related to my work: modelling of the ocean, especially dynamics and everything impacting water level. I took it from the book by Savchenko et al., in the section on deformations.

Problem statement

Assuming that compressibility of water is $\alpha=5\times 10^{-10}\; {\rm Pa^{-1}}$ and the mean ocean depth is around 3.5 km, answer the following questions:

1. Estimate the change in water level if water becomes incompressible.
2. At some places the ocean is 10 km deep. Estimate the change of water density between the 10 km depth and the surface.
3. What is the deformation energy stored in 1 ${\rm m^3}$ of water at the 10 km depth?

Solution

It is probably easy to show that the weight of atmosphere is negligible in comparison with the weight of water pushing and compressing the ocean. Therefore I will neglect the impacts of atmospheric pressure for this problem and will assume that the only cause for the compression of the ocean is its weight.

First, let's remember the definition of the compressibility coefficient, i.e. relative change of volume per unit change of pressure:

\begin{align} \frac{dV}{V} = -\alpha \cdot dP \end{align}

where $V, P$ are the volume and pressure respectively.

To answer the above questions we would need to know how the density of water changes with depth, i.e. $\rho(z)$. We select a layer at depth $z$, and if we push it down by $dz$, the pressure would increase by $dP = \rho(z)\cdot g dz$, since we would have additional amount of water pushing down on our layer. This same pressure increase can be expressed using the definition of compressibility coefficient:

$$ dP = \rho(z)\cdot g dz = \frac{dV}{\alpha V} $$

Then using the relation $V(z) = \frac{m}{\rho(z)}$ and the fact that the mass does not vary during compression, we can express the relative volume change as:

$$ \frac{dV}{V} = \frac{d(m/\rho(z))}{m/\rho(z)} = \frac{d(1/\rho(z))}{1/\rho(z)} \Rightarrow \\ dP = \rho(z)\cdot g dz = -\frac{d(1/\rho(z))}{1/\rho(z) \cdot \alpha} = -\frac{\rho(z)}{\alpha}d\left( \frac{1}{\rho(z)} \right) $$

Then simplifying we get the following equation for the function $\rho(z)$:

$$ d\left( \frac{1}{\rho(z)} \right) = -\alpha g dz $$

Integrating and denoting surface water density as $\rho_0$:

$$ \frac{1}{\rho(z)} = -\alpha g z + \frac{1}{\rho_0} \Rightarrow \\ \rho(z) = \frac{1}{-\alpha g z + \frac{1}{\rho_0}} \Rightarrow \\ \rho(z) = \frac{\rho_0}{1 - \alpha \rho_0 g z} $$

It is easy to estimate that the product in the denominator is at most

$$ \alpha \rho_0 g z \lesssim 10^{-10} \cdot 10^3 \cdot 10^4=10^{-3} \ll 1 $$

Therefore we can further simplify the expression for $\rho(z)$:

\begin{align} \rho(z) \approx \rho_0 \left(1 + \alpha \rho_0 g z \right) \end{align}

Q1: Estimate change in water level if water becomes incompressible.

To figure out the change of the water level we divide the ocean into horizontal layers of width $dz$ and then we calculate compression of each layer due to the weight of water above it. Finally, we can find the water level rise by summing the compressions for all layers.

Following the above plan, we can express mass per unit area of the water layer $dz$ as below:

$$ d\mu = \rho(z)dz = \rho_0 (dz + dh) $$

where $dh$ is the increase of the water layer width due to decreased pressure at the surface (i.e. $P(z=0)=0$).

By reorganizing the terms in the equation above we get:

$$ (\rho(z) - \rho_0) dz = \rho_0 dh $$

Now we substitute the expression we obtained previously for $\rho(z)$ :

$$ \frac{\rho(z) - \rho_0}{\rho_0} dz = dh \Rightarrow \\ \alpha\cdot \rho_0 g z dz = dh $$

Then by integrating from the surface ($z=0$) down to the mean water depth ($z=\overline{H}$), we find the total compression of the water layers, if there was no weight pushing on them, which is equivalent to the problem statement condition imposing water incompressibily:

$$ \Delta h = \int\limits_{0}^{\Delta h} dh = \int\limits_{0}^{\overline{H}}\alpha\cdot \rho_0 g z dz = \alpha\cdot \rho_0 g \frac{\overline{H}^2}{2} \Rightarrow \\ \boxed{ \Delta h \approx 5 \cdot 10^{-10} \cdot 10^3 \cdot 10 \cdot \frac{3.5^2\cdot 10^6}{2} \approx 30.6 \;({\rm m})} $$

Finally, we obtain that if water in the ocean becomes incompressible, the water level will rise by about 30 meters.

Q2: At some places the ocean is 10 km deep. Estimate the difference of water density between the 10 km depth and the surface.

The answer to this question directly follows from the expression we derived for $\rho(z)$ in the beginning of the post:

$$ \Delta\rho (z) = \rho(z) - \rho_0 = \alpha \rho_0^2 g z $$

where $\Delta\rho (z)$ is the difference of water density between the depth $z$ and the surface.

If we substitute 10 km depth (i.e. $z_1 = 10^4 \; {\rm m}$):

$$ \boxed{ \Delta\rho (z_1) = \alpha \rho_0^2 g z = 5 \cdot 10^{-10} \cdot 10^6 \cdot 10 \cdot 10^4 = 50\;({\rm kg/m^3}) } $$

Q3: What is the deformation energy stored in 1 ${\rm m^3}$ of water at the 10 km depth?

The deformation energy stored in comressed water can be computed using work needed to compress a water parcel (let's say a unit cube) from its density at the surface to its density at the 10 km depth. It is easy to show that the compression work $dW$ can be expressed as follows:

$$ dW = PdV \Rightarrow W = \int\limits_{V_0}^{V_H} PdV $$

Using the expression for the compression coefficient we can find pressure as a function of volume $P(V)$:

$$ \int \frac{dV}{V} = -\alpha dP \Rightarrow \ln V = -\alpha P + C $$

Using that at the surface $P(z=0) = 0$ (actually it is equal to the atmospheric pressure, but we neglect it compared to the weight of water) and denoting the volume of the water parcel at the surface as $V_0$ we get:

$$ P = -\frac{1}{\alpha} \ln\frac{V}{V_0} $$

Now we plug it into the expression for work $W$:

\begin{align*} W =& \int\limits_{V_0}^{V_H} -\frac{1}{\alpha} \ln\frac{V}{V_0} dV \\ =& -\frac{V_0}{\alpha} \int\limits_{1}^{V_H/V_0} \ln y \cdot dy \\ =& -\frac{V_0}{\alpha} \left( \left. \ln y \right\rvert_{1}^{V_H/V_0} - \frac{V_H}{V_0} + 1 \right) \\ =& -\frac{V_0}{\alpha} \left( \ln \left( \frac{V_H}{V_0}\right) - \frac{V_H}{V_0} + 1 \right) \\ \end{align*}

Using the relation between densities and volumes of the same parcel of water and expressing the density difference calculated in Q2 we get:

$$ \frac{V_H}{V_0} = \frac{\rho_0}{\rho_H} = \frac{\rho_0}{\rho_0 + \alpha \rho_0^2 g H} = \frac{1}{1 + \alpha \rho_0 g H} $$

Then we can plug the volume ratio into the expression for the energy and get the density as follows:

\begin{align*} \omega & = \frac{W}{V_H} \\ & = -\frac{V_0}{\alpha V_H} \left( \ln \left( \frac{V_H}{V_0}\right) - \frac{V_H}{V_0} + 1 \right) \\ & = -\frac{1 + \alpha \rho_0 g H}{\alpha} \left( -\ln \left( 1 + \alpha \rho_0 g H\right) - \frac{1}{1 + \alpha \rho_0 g H} + 1 \right) \\ \end{align*}

Then using that the $\alpha \rho_0 g H$ is much smaller than 1 as was shown above, we can do the following approximations to the second order of $\alpha \rho_0 g H$:

\begin{align*} \omega & \approx -\frac{1 + \alpha \rho_0 g H}{\alpha} \left( -\alpha \rho_0 g H +\frac{\left(\alpha \rho_0 g H\right)^2}{2} - (1 - \alpha \rho_0 g H + \left(\alpha \rho_0 g H\right)^2) + 1 \right) \\ & = -\frac{1 + \alpha \rho_0 g H}{\alpha} \left( -\frac{\left(\alpha \rho_0 g H\right)^2}{2}\right)\\ & \approx \frac{\alpha}{2}\left( \rho_0 g H\right)^2 \end{align*}

Therefore, the deformation energy density of water at the depth $H$ can be approximated as follows:

$$ \boxed{ \omega = \frac{\alpha}{2}\left( \rho_0 g H\right)^2 } $$

Now, substituting values we get that at the depth $H=10 {\rm km}$:

$$ \omega \approx 0.5 \cdot 5 \cdot 10^{-10}\cdot 10^6 \cdot 10^2 \cdot 10^8=\underline{2.5\times10^6\;({\rm J/m^3})} $$