Ecological Archives E082-016-A3

Douglas C. Speirs and William S. C. Gurney. 2001. Population persistence in rivers and estuaries. Ecology 82:1219-1237.

 

Appendix C. Flow in a tidal river.

 

We seek a solution of the equations describing the dynamics of sea-surface elevation ($h$) and x-velocity ($V_x$)

\begin{displaymath}
\frac{\partial V_x}{\partial t}
=
-g \frac{\partial h}{...
...\partial t}
=
- \int_0^D{\frac{\partial V_x}{\partial x}dz}.
\end{displaymath} (C.1)

 

Our chosen solution must satisfy the following flow boundary conditions at the landward ($x=0$) and seaward ($x=L$) ends of the system

\begin{displaymath}
V_x(0,0,t)=V_R,
\ \ \ \ \ \ \ \ \
V_x(L,0,t)
=V_R+V_T \cos 2 \pi \frac{t}{T} \ \ \ \ \ \ \ \ \ \ \forall \ t ,
\end{displaymath} (C.2)

 

and the conditions of zero wind-stress at the surface and zero slip at the bottom, i.e.

\begin{displaymath}
\frac{\partial V_x}{\partial z}=0
\ \ \textrm{at }z=0,
\...
... \ \ \ \ \ \
V_x(x,D,t)=0 \ \ \ \ \ \ \ \ \ \ \forall \ x,t .
\end{displaymath} (C.3)

 

The problem defined by Eqs. (C.1) - (C.3) is completely linear, so we may expect the solution to be a superposition of the flow generated by the river ($V_s$) and the flow generated by the tide ($V_d$). Since the river input is constant, we expect the river generated flow to be steady (independent of $t$) and uniform (independent of $x$), so we look for a solution of the form

\begin{displaymath}
V_x=V_s(z) + V_d(x,z,t).
\end{displaymath} (C.4)

 

If we associate a surface elevation $h_s$ with the river flow and $h_d$ with the tidal flow, then the assumption that the river generated flow is steady and uniform implies that $\partial h_s/\partial t =0$ and $\partial
h_s/\partial x = -H$, where $H$ is a constant. This is consistent with Eqs. (C.1) if

\begin{displaymath}
\frac{d^2V _s}{dz^2}
= \frac{g}{\Phi_e} \frac{d h_s}{d x } = -\frac{gH}{\Phi_e}
\end{displaymath} (C.5)

 

and

\begin{displaymath}
\frac{\partial V_d}{\partial t}
=
-g \frac{\partial h_d...
...ial t }
=
- \int_0^{D}{\frac{\partial V_d}{\partial x }dz }.
\end{displaymath} (C.6)

 

The general solution of Eq. (C.5) is $V_s=A+Bz -(gH/2\Phi_e)z^2$, where $A$ and $B$ are arbitrary constants. To obtain a zero gradient at $z=0$ (required by equation C.3) we set B=0. To ensure that $V_s(0)=V_R$ in partial fulfilment of the requirements of Eq. (C.2) we set $A=V_R$. Finally we set $H=2\Phi_e
V_R/(gD^2)$, thus matching the requirements of the second element of Eq. (C.3). Our final solution is

\begin{displaymath}
V_s
= V_R\left( 1- \left[ \frac{z}{D}\right]^2 \right).
\end{displaymath} (C.7)

 

To solve Eq. (C.6) we follow Chen et al. (1997) and assume that the solution takes the general form

\begin{displaymath}
V_d = \theta(x,t)\phi(z).
\end{displaymath} (C.8)

 

Substituting this form into Eqs. (C.6) yields

\begin{displaymath}
\phi\frac{\partial \theta}{\partial t}
=
-g \frac{\partia...
...al t }
=
-D \bar{\phi}\frac{\partial \theta}{\partial x },
\end{displaymath} (C.9)

 

where $\bar{\phi}$ represents the average value of $\phi$ over the water column. Differentiating the first element of Eq. (C.9) with respect to time and back substituting for $\partial h_d/\partial t $ yields

\begin{displaymath}
\Phi_e \frac{\partial \theta}{\partial t }
\frac{d^2 \phi...
...\phi
-
gD \frac{\partial^2 \theta}{\partial x^2} \bar{\phi}.
\end{displaymath} (C.10)

 

Now, as a trial, suppose that

\begin{displaymath}
\theta = e^{i \omega t}e^{ikx}
\ \ \ \ \ \ \ \ \textrm{where} \ \ \ \ \ \ \
k^2 = \frac{\omega^2}{gD}
\end{displaymath} (C.11)

 

so that (C.10) can be rewritten

\begin{displaymath}
\frac{d^2}{dz^2}\left(\phi - \bar{\phi} \right)
=
i \frac{\omega}{\Phi_T} \left( \phi - \bar{\phi} \right).
\end{displaymath} (C.12)

 

This has solutions of the form

\begin{displaymath}
\phi-\bar{\phi}=e^{imz}
\ \ \ \ \ \ \ \ \ \ \ \textrm{where} \ \ \ \ \ \ \
m^2 = -i \frac{\omega}{\Phi_e},
\end{displaymath} (C.13)

 

so our generic solution for $v_d$ becomes a linear combination of terms which, without loss of generality, we can write as

\begin{displaymath}
V_d = J_0 e^{\pm i\omega t }e^{\pm ik x }
\left( 1+J_1 e^{\pm m z } \right).
\end{displaymath} (C.14)

 

To ensure that our complete solution matches the boundary conditions (C.2) and (C.3) we select the following combination

\begin{displaymath}
V_d = V_T \left( \frac{\sin kx}{\sin kL} \right)
\Re \left...
...D } \right)
\exp \left( i2\pi \frac{t}{T} \right)
\right\}.
\end{displaymath} (C.15)

 

Thus our final solution for the x-component of scaled velocity becomes

\begin{displaymath}
V_x = V_R\left( 1- \left[\frac{z}{D}\right]^2 \right)
+
...
... mD} \right)
\exp \left( i2\pi \frac{t}{T} \right)
\right\}
\end{displaymath} (C.16)

 

where

\begin{displaymath}
k =\frac{2 \pi}{T}\frac{1}{\sqrt{gD}},
\ \ \ \ \ \textrm{a...
... \ \ \
m =\frac{(1-i)}{\sqrt{2}}\sqrt{\frac{2\pi}{T\Phi_e}}.
\end{displaymath} (C.17)

 

To obtain the z-component of velocity we note that $V_z(x,D,t)=0$ and water is incompressible, so

\begin{displaymath}
V_z(x,z,t) = -\int_D^z \frac{\partial V_x}{\partial x}dz.
\end{displaymath} (C.18)

 

Back substituting from equation (C.16) then gives

\begin{displaymath}
V_z = V_T\left(\frac{k \cos kx}{\sin kL} \right)
\Re \left...
...ft[ D-z+\frac{\sin mz - \sin mD}{m \cos mD} \right]
\right\}.
\end{displaymath} (C.19)


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