Subfield: Thermodynamics

Mid-ocean ridge

Consider a large vessel of water that is situated in a uniform gravitational field with free-fall acceleration $g$. Two vertical rectangular plates parallel to each other are fitted into the vessel so that the vertical edges of the plates are in a tight gap-less contact with the vertical walls of the vessel. Length $h$ of each plate is immersed in water (Fig. 1). The width of the plates along the $y$-axis is $w$, water density is $\rho_{0}$.

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A: without oil

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B: with oil

Figure 1. Parallel plates in water.

Oil of density $\rho_{\text {oil }}\left(\rho_{\text {oil }}<\rho_{0}\right)$ is poured into the space between the plates until the lower level of the oil has reached the lower edges of the plates. Assume that plates and vessel edges are high enough for oil not to overflow them. Surface tension and mixing of fluids can be neglected.

The effects of the surface curvature of the planets can be neglected. You might need to use the formula

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x \text {, when }|x| \ll 1 \text {. }
$$
Context question:
A.1 What is the $x$-component of the net force $F_{x}$ acting on the right plate (magnitude and direction)?
Context answer:
\boxed{$F_{x}=\left(\frac{\rho_{0}}{\rho_{\mathrm{oil}}}-1\right) \frac{\rho_{0} g h^{2} w}{2}$}


Extra Supplementary Reading Materials:

Fig. 2 shows a cross-section of a mid-ocean ridge. It consists of overlaying layers of mantle, crust and ocean water. The mantle is composed of rocks that we assume can flow in geological timescales and therefore, in this problem will be treated as a fluid. The thickness of the crust is much smaller than the characteristic length scale in the $x$-direction, hence, the crust behaves as a freely bendable plate. To high accuracy, such a ridge can be modeled as a two-dimensional system, without any variation of variables along the $y$-axis, which is perpendicular to the plane of Fig. 2. Assume that the ridge length $L$ along the $y$-axis is much larger than any other length introduced in this problem.

At the centre of the ridge the thickness of the crust is assumed to be zero. As the horizontal distance $x$ from the centre increases, the crust gets thicker and approaches a constant thickness $D$ as $x \rightarrow \infty$. Correspondingly, the ocean floor subsides by a vertical height $h$ below the top of the ridge $\mathrm{O}$, which we define as the origin of our coordinate system (see Fig. 2). Water density $\rho_{0}$ and temperature $T_{0}$ can be assumed to be constant in space and time. The same can be assumed for mantle density $\rho_{1}$ and its temperature $T_{1}$. The temperature of the crust $T$ is also constant in time but can depend on position.



It is known that, to high accuracy, the crustal material expands linearly with temperature $T$. Since water and mantle temperatures are assumed to be constant, it is convenient to use a rescaled version of the thermal expansion coefficient. Then $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$, where $l$ is the length of a piece of crustal material, $l_{1}$ is its length at temperature $T_{1}$, and $k_{l}$ is the rescaled thermal expansion coefficient, which can be assumed to be constant.

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Figure 2. Mid-ocean ridge. Note that the $z$-axis is pointing downwards.

Question: A.2 Assuming that the crust is isotropic, find how its density $\rho$ depends on its temperature $T$. Assuming that $\left|k_{l}\right| \ll 1$, write your answer in the approximate form

$$
\rho(T) \approx \rho_{1}\left[1+k \frac{T_{1}-T}{T_{1}-T_{0}}\right]
$$

where terms of order $k_{l}^{2}$ and higher are neglected. Then, identify constant $k$.

Solution:
Consider a rectangular mass element of the crust. Since relation $l(T)=l_{1}\left[1-k_{l}\left(T_{1}-T\right) /\left(T_{1}-T_{0}\right)\right]$ holds for all three dimensions of the solid, its volume $V$ satisfies

$$
V=V_{1}\left(1-k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)^{3}
$$

where $V_{1}$ is the volume at $T=T_{1}$. If the mass of the element is $m$, density is then

$$
\rho(T)=\frac{m}{V}=\frac{m}{V_{1}}\left(1-k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)^{-3}=\rho_{1}\left(1-k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)^{-3}
$$



Since $k_{l} \ll 1$, this can be approximated as

$$
\rho(T) \approx \rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)
$$

so that $k=3 k_{l}$.
A.2

$\rho(T) \approx \rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)$.

$k=3 k_{l}$.

Final answer: $\rho_{1}\left(1+3 k_{l} \frac{T_{1}-T}{T_{1}-T_{0}}\right)$ , $k=3 k_{l}$