Energetics of Burrow Excavation in Sediment

We now consider an alternate estimate of energy associated with burrowing. Consider a cylindrical burrow perpendicular to the sediment–water interface in which the sediment is excavated from the substrate and redeposited at the sediment–water interface. At a minimum, burrow excavation does work against gravity by lifting particles toward the sediment–water interface:
$$\begin{array}{c}\hfill E_{\mathit{gr}}=mgz,\end{array}$$
where E_gr is the change in gravitational energy [J], m is the mass of sediment moved [M], g is the gravitational acceleration [L] [t⁻²], and z is the vertical distance [L]. The (buoyant?) mass of some portion of the cylindrical burrow with a vertical thickness of Δz is
$$\begin{array}{c}\hfill m=(1-\varphi )({\rho }_s-{\rho }_w)\pi b^2\mathrm{\Delta }z,\end{array}$$
where ϕ is the porosity of the sediment, ρ_s and ρ_w are the densities of sediment and water, respectively, b is the burrow radius, and z is the distance from the center-of-mass to the sediment–water interface. The work of lifting n segments is
$$\begin{array}{c}\hfill \sum _{i=1}^n{E}_{\mathit{gr}}=\sum _{i=1}^n\left[(1-\varphi )({\rho }_s-{\rho }_w)\pi b^2\mathrm{\Delta }z\right]g{z}_i.\end{array}$$
Taking the limit as n goes to infinity and Δz goes to zero, the work for excavating a cylindrical burrow that extends to some depth d below the sediment–water interface is
$$\begin{array}{c}\hfill \begin{array}{cc}\hfill E_{\mathit{gr}}& ={\int }_0^d\left[(1-\varphi )({\rho }_s-{\rho }_w)\pi b^{2}dz\right]gz\hfill \\ \hfill & =\frac{1}{2}(1-\varphi )({\rho }_s-{\rho }_w)\pi b^{2}g{d}^2.\hfill \end{array}\end{array}$$
The result in Eq. 19 is the same as if we had used the entire mass of the burrow in Eq. 17 and raised the center-of-mass from half the burrow depth to the surface.
We now consider the work done in excavating n identical burrows:
$$\begin{array}{c}\hfill E_{\mathit{gr}}=n\left[\frac{1}{2}(1-\varphi )({\rho }_s-{\rho }_w)\pi b^{2}g{d}^2\right]\end{array}$$
If we approximate the rate at which new burrows are created as a continuous process, then
$$\begin{array}{c}\hfill \frac{E_{\mathit{gr}}}{\mathit{dt}}=\frac{\mathit{dn}}{\mathit{dt}}\left[\frac{1}{2}(1-\varphi )({\rho }_s-{\rho }_w)\pi b^{2}g{d}^2\right].\end{array}$$
Under this approximation, the rate at which new burrows are created is related to the linear burrowing velocity by
$$\begin{array}{c}\hfill \frac{\mathit{dn}}{\mathit{dt}}=\frac{1}{d}\frac{\mathit{dz}}{\mathit{dt}}.\end{array}$$
Using Eq. 14, the relationship to the volumetric burrowing rate reported by (22 ) is
$$\begin{array}{c}\hfill \frac{\mathit{dn}}{\mathit{dt}}=\frac{R}{\pi b^{2}d}.\end{array}$$
Substituting Eq. 23 into Eq. 21, we arrive at
$$\begin{array}{c}\hfill \frac{d{E}_{\mathit{gr}}}{\mathit{dt}}=\frac{1}{2}(1-\varphi )({\rho }_s-{\rho }_w)Rgd.\end{array}$$
Dimensional analysis shows that Eq. 24 has units of [J] [L⁻²] [t⁻¹]. The interpretation of these units is the same as for Eq. 15 in the preceding section: a rate of energy transfer per unit area of the sea floor.