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Constant in/out flow, linear level dependent estuary.

If we have a constant river+rainfall+evaporation inflow, one square sided cell and a linearly level dependent estuary, (See figure 15)

  figure746
Figure 15: Single lake cell the level dependent estuary and time dependent inflow


equation1296

Opening or closing Potter's channel would correspond to changing the value of k.

Rearranging slightly so we may more easily see the homogeneous part.
displaymath3317

Then the solution is...
equation1298

Integrating this gives us...
 equation1300

Note one small detail. tex2html_wrap_inline3319 is not the volume of water in the cell at time t=0. tex2html_wrap_inline3319 is merely a constant of integration. Now for interests sake let us put this in terms of tex2html_wrap_inline3197, the volume of water at time t=0.
equation1302

Which implies...
equation1304

So the equation (51) for the volume becomes...

Volume as a function of time.


 equation1325

A pair of terms of the form tex2html_wrap_inline3329 is the mathematics way of taking the system from the initial condition A at time t=0 to the final condition B at time tex2html_wrap_inline3337.

The final steady state condition tex2html_wrap_inline3339 is made up of two terms...

  1. Term one is a head difference due to in flow I. The magnitude of the head difference due to I is governed by the estuary flow constant k.
  2. The last term is just the effect of the sea level, which 'pins' the differential equation to move about sea level. Without a level dependent estuary the volumes can wander to arbitrarily large or small values.

Denote the salt load by S. Now the load calculation, assuming no salt in the inflow, is...


 eqnarray1327
Where C is the salt concentration of sea water (tex2html_wrap_inline3343) when V/A > L and is S/V when the flow is from lake to sea.


Next: Sea to lake flow. Up: Potter's channel type problems. Previous: Potter's channel type problems.

John Carter
Tue Jun 17 09:50:07 SAT 1997