Analytical solution for the tidally induced Lagrangian residual current in a narrow bay

Abstract

In a weakly nonlinear tidal system, the depth-averaged equations for the first-order Lagrangian residual velocity (LRV) are deduced systematically. For the case of a narrow bay, the equations are solved analytically and the results for a specific bottom profile are discussed in detail. According to the pattern of the first-order LRV, the bay can be divided into three parts, namely an inner part, a transitional zone, and an outer part. For the given depth profile, the streamline of the first-order LRV for a shorter bay is a part of that for a longer bay. The first-order LRV depends on a nondimensional parameter that combines the influences of the bottom friction coefficient, the tidal period and the averaged water depth. The form of the bottom friction also has a significant influence on the first-order LRV. The second-order LRV, i.e., the Lagrangian drift, is analytically solved and shows dependence on the initial tidal phase. The LRV differs from the Eulerian residual transport velocity both quantitatively and qualitatively. It is demonstrated that the residual currents obtained according to other definitions may cause misunderstanding of the mass transport in water exchange applications.

Appendix: The derivation of equations for uL

By applying the operator (Eq. 26) to Eq. 18 and making use of the definition of u _E in Eq. 3, one gets

$$ \frac{\partial hu_E}{\partial x}+\frac{\partial h\upsilon_E}{\partial y}= -\left( \frac{\partial \langle u_0 \zeta_0 \rangle}{\partial x} +\frac{\partial \langle \upsilon_0 \zeta_0 \rangle}{\partial y} \right). $$

(52)

The substitution of Eqs. 23–25 into Eq. 4 gives

$$ hu_S=\langle u_0\zeta_0 \rangle+\frac{\partial}{\partial y} \langle h u_0\eta_0 \rangle, $$

(53)

$$ h\upsilon_{S}=\langle \upsilon_0\zeta_0 \rangle+\frac{\partial}{\partial x} \langle h \upsilon_0\xi_0 \rangle. $$

(54)

By performing $\partial / \partial x$ (Eq. 53) + $\partial / \partial y$ (Eq. 54) and taking into consideration of Eq. 52, the continuity equation of u _L , Eq. 27, can be obtained.

The momentum equations of the LRV are derived as follows. By applying Eq. 26 to Eq. 19 and making use of Eq. 2, one gets

$$\begin{array}{rll} 0=&-&\frac{\partial \zeta_E}{\partial x} -\beta_T \frac{u_L}{h} +\beta_{T1} \frac{\langle u_0 \zeta_0 \rangle}{h^2}\\ &-&\left\langle u_0\frac{\partial u_0}{\partial x} + \upsilon_0 \frac{\partial u_0}{\partial y}\right\rangle +\beta_T \frac{1}{h} \left\langle \xi_0 \frac{\partial u_0}{\partial x } + \eta_0 \frac{\partial u_0}{\partial y }\right\rangle. \end{array}$$

(55)

By applying the operator $(\xi_0 \partial/\partial x+\eta_0 \partial/\partial y)$ to Eq. 14 and making use of Eq. 15, followed by applying Eq. 26, one yields

$$\begin{array}{lll} \left\langle \xi_0\frac{\partial^2 u_0}{\partial x \partial t}+ \eta_0\frac{\partial^2 u_0}{\partial y \partial t} \right \rangle= -\left\langle \xi_0\frac{\partial^2 \zeta_0}{\partial x^2} \right \rangle\\ \,\quad-\beta_T \left\langle \xi_0 \frac{\partial}{\partial x}\left (\frac{u_0}{h}\right)+ \eta_0 \frac{\partial}{\partial y}\left(\frac{u_0}{h}\right) \right\rangle. \end{array}$$

(56)

The left side of Eq. 56 is rearranged with integration by parts. Furthermore, by making use of Eq. 6 and the periodicity of the variables, one gets

$$\begin{array}{rll} \left\langle \xi_0 \frac{\partial^2 u_0}{\partial x \partial t}+\eta_0 \frac{\partial^2 u_0}{\partial y \partial t} \right \rangle&=& \left \langle \frac{\partial}{\partial t} \left(\xi_0 \frac{\partial u_0}{\partial x} + \eta_0 \frac{\partial u_0}{\partial y} \right) \right \rangle \\ && - \left \langle \frac{\partial \xi_0}{\partial t} \frac{\partial u_0}{\partial x} +\frac{\partial \eta_0}{\partial t} \frac{\partial u_0}{\partial y} \right \rangle\\ &=& -\left \langle u_0 \frac{\partial u_0}{\partial x}+\upsilon_0 \frac{\partial u_0}{\partial y} \right \rangle. \end{array}$$

(57)

Then by changing Eq. 56 to the following form

$$\begin{array}{rll} -\left\langle\! u_0 \frac{\partial u_0}{\partial x} \!+\!\upsilon_0 \frac{\partial \upsilon_0}{\partial y} \!\right \rangle= &-&\!\left \langle \xi_0\frac{\partial^2 \zeta_0}{\partial x^2} \right \rangle\\ &-& \beta_T \frac{1}{h} \left \langle \xi_0 \frac{\partial u_0}{\partial x}+ \eta_0 \frac{\partial u_0}{\partial y} \right \rangle \\ &-& \beta_T \left \langle\! \xi_0 u_0\frac{\partial}{\partial x}\left( \frac{1}{h}\right)\!+\! \eta_0 u_0 \frac{\partial}{\partial y}\left( \frac{1}{h}\right) \!\right \rangle, \end{array}$$

and by adding Eq. 55 to the above equation, one gets

$$\begin{array}{rll} 0= &-&\frac{\partial \zeta_E}{\partial x}- \beta_T \frac{u_L}{h} +\beta_{T1} \frac{\langle u_0 \zeta_0 \rangle}{h^2} -\left \langle \xi_0\frac{\partial^2 \zeta_0}{\partial x^2} \right \rangle \\ &-&\beta_T \left \langle \xi_0 u_0\frac{\partial}{\partial x}\left( \frac{1}{h}\right)+ \eta_0 u_0 \frac{\partial}{\partial y}\left( \frac{1}{h}\right) \right \rangle. \end{array}$$

(58)

By applying $\xi_0\partial/\partial x$ to Eq. 14, one obtains

$$\xi_0 \frac{\partial ^2 \zeta_0}{\partial x^2}= -\xi_0 \left( \frac{\partial ^2 u_0}{\partial x\partial t}+\beta_T \frac{\partial}{\partial x }\left(\frac{u_0}{h} \right )\right ), $$

(59)

and then $\partial/\partial x$ [ξ ₀ · (Eq. 14)] leads to

$$ \frac{\partial}{\partial x} \left ( \xi_0 \frac{\partial \zeta_0}{\partial x} \right )= -\frac{\partial}{\partial x} \left (\xi_0 \frac{\partial u_0}{\partial t} +\beta_T \xi_0 \frac{u_0}{h} \right). $$

(60)

Then, Eq. 59 is multiplied with 2 and subtracted with Eq. 60; and the resulting equation is averaged over one tidal period. By making use of Eq. 6 and the periodicity of the zeroth-order solution, one obtains

$$ \left \langle \xi_0 \frac{\partial ^2 \zeta_0}{\partial x^2} \right \rangle= \left \langle \frac{1}{2} \frac{\partial} {\partial x} \left( \xi_0 \frac{\partial \zeta_0}{\partial x}\right ) \right \rangle+\beta_T \frac{1}{h} \left \langle u_0 \frac{\partial \xi_0}{\partial x} \right \rangle. $$

(61)

Substituting Eq. 61 into Eq. 58 leads to

$$\begin{array}{rll}0= &-&\frac{\partial \zeta_E}{\partial x}-\beta_T \frac{u_L}{h} +\beta_{T1} \frac{\left \langle u_0 \zeta_0 \right \rangle}{h^2}-\left \langle \frac{1}{2} \frac{\partial} {\partial x} \left( \xi_0 \frac{\partial \zeta_0}{\partial x} \right ) \right \rangle\nonumber \\ &-&\beta_T \left \langle \xi_0 u_0 \frac{\partial }{\partial x} \left(\frac{1}{h} \right ) + \eta_0 u_0 \frac{\partial }{\partial y} \left(\frac{1}{h} \right ) \right \rangle\nonumber \\ &-&\beta_T \frac{1}{h} \left \langle u_0 \frac{\partial \xi_0}{\partial x} \right \rangle. \end{array}$$

(62)

For the case of single tidal frequency $\langle \xi_0 u_0 \rangle =\langle \eta_0 \upsilon_0 \rangle=0$, thus

$$\begin{array}{lll} -\beta_T \left \langle \xi_0 u_0 \frac{\partial }{\partial x} \left(\frac{1}{h} \right ) + \eta_0 u_0 \frac{\partial }{\partial y} \left(\frac{1}{h} \right ) \right \rangle= \beta_T \frac{1}{h^2} \frac{\partial }{\partial y} \langle h u_0 \eta_0 \rangle\\ \quad+ \beta_T \frac{1}{h} \left \langle \xi_0 \frac{\partial \upsilon_0}{\partial y} -\eta_0 \frac{\partial u_0}{\partial y} \right \rangle,\\ -\beta_T \frac{1}{h} \left \langle u_0 \frac{\partial \xi_0}{\partial x}\right \rangle= \beta_T \frac{1}{h} \left \langle \xi_0 \frac{\partial u_0}{\partial x} \right \rangle. \end{array}$$

By applying the above results to Eq. 62, Eq. 28 can be obtained. It is trivial to obtain the momentum equation in the y direction.