Influence of the back-barrier basin length on the geometry of ebb-tidal deltas

Abstract

The characteristics of ebb-tidal deltas are determined by the local hydrodynamics. The latter depend, among others, on the geometry of the adjacent back-barrier basin. Therefore, interventions in the back-barrier basin can affect the geometry of ebb-tidal deltas. In this study, the effect of the length of the back-barrier basin on the sand volume and spatial symmetry of ebb-tidal deltas is quantified with the use of a numerical model. It is found that the length of the back-barrier basin affects the tidal prism, the amplitude and phase of the primary tide and its overtides, and the residual currents that, together, determine the sand volume of the ebb-tidal delta. In particular, it is found that no unique relationship exists between tidal prism and sand volume of an ebb-tidal delta. The spatial symmetry of ebb-tidal deltas is also found to be affected by the length of the back-barrier basin. This is because the basin length determines the phase difference between alongshore and cross-shore tidal currents. The numerical model results give a possible explanation for the changes that are observed in the geometry of the ebb-tidal deltas that are located seaward of the Texel Inlet and Vlie Inlet after the closure of the Zuiderzee.

Appendix: A One-dimensional analytical model

To gain insight into the effect of the length of a back-barrier basin on the hydrodynamics in a tidal inlet, a simple model was analyzed, which consists of the linearized shallow water equations for a configuration of two connected one-dimensional channels with a constant width, W _n , and depth, H _n (Fig. 10). The equations read

$$\begin{array}{@{}rcl@{}} \frac{\partial \tilde{v}_{n}}{\partial t} = -g \frac{\partial \tilde{\eta}_{n}}{\partial y} - \frac{\lambda_{n}}{H_{n}} \tilde{v}_{n}, \end{array} $$

(A.1)

$$\begin{array}{@{}rcl@{}} \frac{\partial \tilde{\eta}_{n}}{\partial t} + H_{n} \frac{\partial \tilde{v}_{n}}{\partial y} = 0, \end{array} $$

(A.2)

where n = 1 represents the tidal inlet and n = 2 the back-barrier basin, $\tilde {v}$ is the along channel velocity, g is the acceleration due to gravity, $\tilde {\eta }$ is the sea surface height, y is the horizontal coordinate, t is the time, and λ _n is a linearized friction coefficient (Eqs. A.15 and A.16). These equations are solved by applying the following boundary conditions:

$$ \tilde{\eta}_{1} = \mathbb{R}\left(\hat{Z}e^{-i \sigma t}\right), {\kern4pc} \text{at } y = -L_{1}, \\ $$

(A.3)

$$ \tilde{\eta}_{1} = \tilde{\eta}_{2}, {\kern6.8pc} \text{at } y = 0,\\ $$

(A.4)

$$ W_{1}H_{1}\tilde{v}_{1} = W_{2}H_{2}\tilde{v}_{2}, {\kern2.7pc} \text{at } y = 0,\\ $$

(A.5)

$$ \tilde{v}_{2} = 0, {\kern7.2pc} \text{at } y = L_{2}. $$

(A.6)

These boundary conditions imply that at y=−L ₁, $\tilde {\eta }_{1}$ is forced by a harmonic variation in sea surface height; at y=0 there are two boundary conditions, viz. conservations of mass and continuity of $\tilde {\eta }$. Finally, at y=L ₂ the velocity $\tilde {v}_{2}$ vanishes. Solutions for $\tilde {\eta }_{n}(y,t)$ and $\tilde {v}_{n}(y,t)$ that satisfies Eqs. A.1 and A.2 are of the type

$$ \tilde{\eta}_{n}(y,t) = \mathbb{R}\left(N_{n}(y)e^{-i\sigma t}\right), \\ $$

(A.7)

$$ \tilde{v}_{n}(y,t) = \mathbb{R}\left(V_{n}(y)e^{-i\sigma t}\right). $$

(A.8)

Following standard methods and using Eqs. A.3–A.6 the following solutions are obtained:

$$ N_{1}(y) = \hat{Z} \frac{ \chi_{1}\cos(\kappa_{1} y)\cos(\kappa_{2} L_{2}) + \chi_{2}\sin(\kappa_{1} y)\sin(\kappa_{2} L_{2}) }{\chi_{1}\cos(\kappa_{1} L_{1})\cos(\kappa_{2} L_{2}) - \chi_{2}\sin(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2})} ,\\ $$

(A.9)

$$ N_{2}(y) = \frac{1}{2}\hat{Z} \frac{\chi_{1} \left( e^{-i\kappa_{2}(y-L_{2})} + e^{i \left(\kappa_{1} y - \kappa_{2} L_{2}\right)}\right)}{\chi_{1}\cos(\kappa_{1} L_{1})\cos(\kappa_{2} L_{2}) - \chi_{2}\sin(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2})} ,\\ $$

(A.10)

$$\begin{array}{@{}rcl@{}} V_{1}(y) &=& \left(\frac{g\hat{Z} \kappa_{1} H_{1}}{\lambda_{1} - i \sigma H_{1}} \right) \\&&\times \frac{ \chi_{1}\sin(\kappa_{1} y)\cos(\kappa_{2} L_{2}) - \chi_{2}\cos(\kappa_{1} y)\sin(\kappa_{2} L_{2}) }{\chi_{1}\cos(\kappa_{1} L_{1})\cos(\kappa_{2} L_{2}) - \chi_{2}\sin(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2})} ,\\ \end{array} $$

(A.11)

$$\begin{array}{@{}rcl@{}} V_{2}(y)& =& \left(\frac{g\hat{Z} \kappa_{2} H_{2}}{i \sigma H_{2} -\lambda_{2}} \right)\\ &&\times\frac{ \chi_{1}\sin{(\kappa_{2} (L_{2} -y))}}{\chi_{1}\cos(\kappa_{1} L_{1})\cos(\kappa_{2} L_{2}) - \chi_{2}\sin(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2})}.\\ \end{array} $$

(A.12)

Here κ _n is the frictional wave number,

$$ \kappa_{n} = \sqrt{\frac{H_{n} \sigma^{2} + i\sigma\lambda_{n}}{g {{H^{2}_{n}}}}}, $$

(A.13)

and

$$ \chi_{n} = \frac{H_{n} W_{n} \kappa_{n}}{\left(\lambda_{n} / H_{n}\right) - i\sigma}. $$

(A.14)

In order to establish that the linearized bottom stress yields the correct tidally averaged dissipation of energy, the method of Lorentz (1926) (Zimmerman 1992) is applied. With this method λ _n is determined recursively. New solutions for V _n (x) and N _n (x) are calculated until the values for λ _n that were used to obtain the solutions are within 0.5 % of the $\lambda _{n}^{*}$ that would give the correct tidally averaged dissipation of energy, i.e.:

$$ \lambda_{n}^{*} = \frac{8 c_{\text{D}}}{3\pi}\frac{{\int}^{L_{n}} |V_{n}(y,\lambda_{n})|^{3}dy}{{\int}^{L_{n}} |V_{n}(y,\lambda_{n})|^{2}dy}, \\ $$

(A.15)

$$ \left|\left(\frac{\lambda_{n}^{*}}{\lambda_{n}} - 1\right)\right| < 0.005, $$

(A.16)

where the drag coefficient, c _D, is a constant parameter.

Fig. 10

Domain of the analytical model

Since the tidal prism, P, in this model is given by

$$ P = \frac{1}{2}H_{1}W_{1}{{{\int}_{0}^{T}}}|\tilde{v}_{1}(-L_{1},t)|dt. $$

(A.17)

it follows from Eq. A.11 that

$$ P = \left|\frac{2 g \chi_{1} \hat{Z} }{\sigma} \frac{ \chi_{1}\sin(\kappa_{1}L_{1})\cos(\kappa_{2} L_{2}) + \chi_{2}\cos(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2}) }{\chi_{1}\cos(\kappa_{1} L_{1})\cos(\kappa_{2} L_{2}) - \chi_{2}\sin(\kappa_{1} L_{1})\sin(\kappa_{2} L_{2})}\right|. $$

(A.18)