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 $H=10 {\rm km}$:

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

Minimum work required to bend a metal rod

This problem is from the section on deformations obbeying the linear relation between reaction force and deformation length. This one requires knowledge of Young modulus which allows to express reaction force as a function of relative deformation. In one-dimensional case the law looks as follows:

$$ \sigma = E\varepsilon $$

where $\sigma=F/A$ is the reaction tension (${\rm N/m^2}$) of the material in response to the relative deformation $\varepsilon=\Delta L / L$.

This problem is taken from the book Savchenko et al and it took me some time to understand to my satisfaction, therefore it is written up here.

Problem statement

Determine the minimum amount of work required to bend a rod of length $l$ with square cross section ($a \times a$) into a ring. Young modulus of the material is $E$ and $l \gg a$.

Solution

The approach to take here was hinted to me by the previous problem which asked for the expression of the energy density for a deformed material. It is relatively easy to derive using the above expression for the deformation tension that the energy density per unit volume ($\omega$) can be expressed as follows:

$$ \omega = \frac{dW}{dV} = \frac{E\varepsilon^2}{2} $$

So the problem now is reduced to finding the ring configuration that would minimize the energy (work) due to deformation of the rod:

$$ W = \int\omega dV \Rightarrow W_{\rm min} = \min_{V_{\rm ring}}\left( \int \omega dV \right) $$

In the bent state we can divide the ring into multiple strips of width $dr$ with the smallest having the radius $R$. We assume that the length of the longest (outer) strip would be $R+a$, i.e. here we neglect radial deformations and assume that the width of the rod as well as its height won't change. Then the deformation energy $dW$ stored in each strip would be:

$$ dW = \omega dV = \frac{E\varepsilon^2}{2}dV =\frac{E\varepsilon^2}{2}a\cdot 2\pi r dr $$

Therefore the total energy of the deformed rod would be a sum of energy of each of the strips:

$$ W = \int\limits_{R}^{R+a}\frac{E\varepsilon^2}{2}a\cdot 2\pi r dr $$

Then noting that for each strip the relative deformation $\varepsilon$ is $(2 \pi r -l )/l $, we get: $$ W(R) = \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr $$

Now we need to find the radius of the shortest strip $R$ (which would also define the radii of all the longer strips) that minimizes $W(R)$, i.e.:

$$ W_{\min} = \min_{R} W(R) = \min_{R}\left( \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr \right) $$

Intuitively, I figured that the minimizing configuration would include the undeformed strip (i.e. the one in bent state having the same length $l$ as before). To minimize the deviations of the other strips, the undeformed strip would have to be in the middle, so that leaves us with $R = \frac{l}{2\pi} - \frac{a}{2}$ and once we have this the solution is straightforward. But I still would like to find the minimizing $R$ in a more rigorous way.

If one accepts the above arguments then (If not please see another way of determining $R$ in the note section at the end of the post) the expression for $W_{\min}$ becomes:

$$ W_{\min}=W\left(R=\frac{l}{2\pi} - \frac{a}{2}\right)=\int\limits_{\frac{l}{2\pi} - \frac{a}{2}}^{\frac{l}{2\pi} + \frac{a}{2}}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr $$

Making a substitution $y = 2 \pi r$: $$ W_{\min} = \frac{Ea}{4\pi l^2}\int\limits_{l - \pi a}^{l + \pi a}(y - l)^2\cdot y dy $$

Then simplifying: $$ W_{\min} = \frac{Ea}{4\pi l^2}\left(\int\limits_{\:l - \pi a}^{l + \pi a}(y - l)^3 dy + l\cdot\int\limits_{l - \pi a}^{l + \pi a}(y - l)^2 dy \right) $$

And another substitution for convenience $z=y-l$: $$ W_{\min} = \frac{Ea}{4\pi l^2}\left(\int\limits_{\: - \pi a}^{\pi a}z^3 dz + l\cdot\int\limits_{- \pi a}^{\pi a}z^2 dz \right) $$

Noticing that the first integral above is from an odd function over a symmetric interval with respect to 0, therefore it is 0 (or you could simply integrate to convince yourself) and we get: $$ W_{\min} = \frac{Ea}{4\pi l^2}l\cdot\int\limits_{- \pi a}^{\pi a}z^2 dz = \frac{Ea}{4\pi l^2}l\cdot \frac{2 \pi^3 a^3}{3} $$

Then simplifying we get the final expression for the minimum amount of work needed to bend the rod:

$$ \boxed{W_{\min} = \frac{E\pi^2 a^4}{6 l}} $$

Note on finding the minimum radius of the minimizing configuration of bent strips

The general expression for the deformation energy (work) of the bent rod is:

$$ \min_{R} W(R) = \min_{R}\left( \int\limits_{R}^{R+a}\frac{E(2 \pi r - l)^2}{2l^2}a\cdot 2\pi r dr \right) $$

We need to find the $R_0$ for which $W(R)$ is minimized. The minimizing configuration should contain the undeformed strip and therefore we can limit our search for $R_0$ to the closed interval $\left[ \frac{l}{2\pi} - a, \frac{l}{2\pi} \right ]$.

Reorganizing the terms

$$ W(R) = \min_{R}\left( \frac{Ea}{8\pi l^2} \int\limits_{2\pi R - l}^{2\pi R+ 2\pi a - l}z^2\cdot (z + l) dz \right) $$

Let's change the independent variable from $R$ to $Z = 2\pi R - l$ (where $Z \in [-2\pi a\,,\: 0]$), for convenience:

$$ W(Z) = \min_{Z}\left( \frac{Ea}{8\pi l^2} \int\limits_{Z}^{Z+ 2\pi a}z^2\cdot (z + l) dz \right) $$

Equivalently, we need to determine $Z$, which will minimize the following function:

$$ I(Z) = \int\limits_{Z}^{Z+ 2\pi a}z^2\cdot (z + l) dz $$

To find the extrema of $I(Z)$ we solve for roots of $\frac{dI(Z)}{dZ}=0$.

By differentiation of the integral with variable limits we get the following expression for the derivative:

$$ \frac{dI(Z)}{dZ}=(Z+2\pi a)^2 \cdot (Z+2\pi a + l) - Z^2 \cdot (Z + l)=0 $$

Simplifying we get the following equation for the extrema of $I(Z)$:

$$ Z^2 2 \pi a + 4\pi a Z^2 + 4\pi a Z (2\pi a + l) + 4\pi^2 a^2 Z + 4\pi^2 a^2 (2\pi a + l) = 0 $$

Simplifying further, we get the following quadratic equation with respect to $Z$:

$$ 3Z^2 + 2 Z (3\pi a + l) + 2\pi a (2\pi a + l) = 0 $$

The exact solution is then:

$$ Z_{1,2} = \frac{-(3\pi a + l) \pm \sqrt{l^2 - 3 \pi^2a^2}}{3} $$

then using the condition $l \gg a$ we neglect second order term $3\pi^2(a/l)^2$:

$$ \left [ \begin{array}{l} Z_1 = -\pi a \\ Z_2 = -\pi a - 2l/3 \\ \end{array}\right. $$

Since $\frac{dI(Z)}{dZ}$ is quadratic, and $Z_2 < Z_1$, it is easy to see that

$$ \left\{ \begin{array}{l} \frac{dI(Z)}{dZ} \geq 0\,,\; Z\leq Z_2 \\ \frac{dI(Z)}{dZ} \leq 0\,,\; Z_2 < Z \leq Z_1 \\ \frac{dI(Z)}{dZ} > 0\,,\; Z > Z_1 \\ \end{array}\right. $$

Therefore we can conclude that $Z_1$ is a local mininmum, and $Z_2$ is a local maxinmum of $I(Z)$. Hence $Z_1 = -\pi a$ is a candidate for a minimum of $I(Z), Z\in [-2 \pi a\,,\: 0]$. To make sure that $Z_1$ is the minimum we need to check that $I(Z_1) < I(-2 \pi a)$.

Let's compute directly these values:

$$ I(Z_1) = \int\limits_{-\pi a}^{\pi a}z^2\cdot (z + l) dz = \frac{2}{3}l \pi^3 a^3 $$$$ I(-2 \pi a) = \int\limits_{-2\pi a}^{0}z^2\cdot (z + l) dz = 4 \pi^4 a^4 + \frac{8}{3}l \pi^3 a^3 $$

From the two above expressions it is obvious that $Z=-\pi a$ minimizes $I(Z), Z \in [-2\pi\,,\: 0]$.

Finally, we can find the $R_0$ which minimizes energy of deformation:

$$ Z_0 = -\pi a = 2 \pi R_0 - l \Rightarrow \boxed{R_0 = \frac{l}{2\pi} - \frac{a}{2}} $$

Which is the same as we guessed before.